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Mathematical Theory and Modeling www.iiste.org

ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.10, 2012

21

DERIVATION OF A CLASS OF HYBRID ADAMS MOULTON

METHOD WITH CONTINOUS COEFFICIENTS

Alagbe, Samson Adekola.1 Awogbemi, Clement Adeyeye

2* Amakuro Okuata

3

1, 3 Department of Mathematics, Bayelsa State College of Education, Okpoama, Nigeria.

2. National Mathematical Centre, Sheda-Kwali, P.M.B 118, Abuja, Nigeria.

* E-mail of the corresponding author: [email protected]

Abstract

This research work is focused on the derivation of both the continuous and discrete models of the hybrid

Adams Moulton method for step number k =1 and k = 2. These formulations incorporate both the off –

grid interpolation and off- grid collocation schemes. The convergence analysis reveals that derived

schemes are zero stable, of good order and error constants which by implication shows that the schemes

are consistent.

Keywords: Hybrids Schemes, Adams Methods, Linear K–Step Method, Consistency, Zero Stable

1.0 Introduction

The derivation of hybrid Adams Moulton Schemes of both the continuous and discrete forms for the off-

grid interpolation and the off – grid collocation system of polynomials is our primary focus in this

research work. Sequel to this would be to ascertain the zero stability of each of the discrete forms.

Performances of these schemes on solving some non stiff initial value problems shall be affirmed in the

second phase of this work which will focus on the application of these derived scheme and comparism of

the discrete schemes with the single Adams-Moulton Methods and its alternative in Awe (1997) and

Alagbe (1999)

In this paper we regard the Linear Multi-step Method and Trapezoidal ( Adams – Moulton Method) as

extrapolation and substitution methods respectively.

We also define k –Step hybrid schemes as follows:

k

i i

vnriniini fhfhky0 0

. …………………………………………………(1.1)

where 00,1 andk are not both zero and v },...1,0{ k and of course ),( vnvnvn yxff .

Gregg & Stetter (1964) satisfied the co-essential condition of zero-stability, the aim of which to reduce

some of the difficult inherent in the LMM (i.e poor stability) was achieved.

mailto:[email protected]



22

Hybrid scheme was as a result of the desire to increase the order without increasing the step number and

then without reducing the stability interval. However, the hybrid methods have not yet gained the

popularity deserved due to the presence of off-grid point which requires special predicator which will not

alternate the accuracy of the corrector to estimate them.

1.1 Adams Methods

This is an important class of linear multi-step method of the form:

)2.1..(.................................720

251

8

3

12

5

2

11 432

1 nnn fhyy

The corrected form of Adams-Moulton form is expressed as:

)3.1..(.................................120

19

24

1

12

1

2

11 1

432

1

nnn fhyy

The trapezoidal scheme is a special case of the Adams –Moulton method, in which only the first two terms

in the bracket is retained. This method is of the highest order among the single –step method. It is being

expressed as

nnn ffh

y 112

2.0 Derivation of Continuous and Discrete Hybrid Adams-Moulton Scheme

Our concern primarily is to derive the continuous and discrete form of both one-step and two-step

Adams-Moulton scheme and consequently carried out the error and zero-stable analysis of the discrete

forms. Matrix inversion technique was the tool implored to achieve the derivation of the continuous form.

2.1 Derivation of Multi-Step Collocation Method

In order to derive the continuous and discrete one and two step hybrid Adams-Moulton schemes, we

employed the approach used by Sirisena (1997) where a k-step multi-step collocation method point was

obtained as:

1.2.......................................................,1

0

1

0

m

j

jjjn

k

j

j xyjfxhxyxxy x

Where xj and xj are the continuous coefficients

We defined xj and xj respectively as:



23

1

0

1,

1

0

1,

2.2...............................................................................................mt

i

i

ijj

mt

i

i

ijj

xhxh

xx

To get xj and xj, we arrived at matrix equation of the form DC=I………… (2.3)

where I is the identity matrix of dimension (t + m) (t + m), D and C are defined as

4.2.........................................

1201

1210

1

.

1

1

2

1

1

11

_2

0

1

0

1

11

2

11

1

11

2

11

12

mt

m

t

mm

mtt

o

mt

kn

t

knknkn

mt

n

t

nnn

mt

n

t

nnn

xxx

xx

mtt

mttx

xxxx

xxxx

xxxx

D

Thus, matrix (2.4) is the multi-step collocation matrix of dimension (t+m) (t+m) while matrix C of the

same dimension whose columns give the continuous coefficients given as:

5.2..........................

,10,110

2,1022,11202

1,1011,11101

jmmmjmjtmjmj

mt

mt

hh

hh

hh

C

We define t as the number of interpolation points and m is the number of collocation points. From (2.3),

we notice that If we define

This implies that a suitable algorithm for getting the elements of C is the following:



24

9.2.........................................................................................................,.........,...2,1;

.,..2,1,,,

8.2........................................................................................................................,...,2,1

,...,2,1.;,...2,1,

7.2...........................................................................,...2,1.,,...2,1,

.,...2,1,

6.2................................................................................,.........,...2,1,

1

1

,

1

1

,

1

1

1

,

,

,

,,

njldu

nijiluldu

nidl

ninjijuldl

njnilelee

njl

ee

njicueC

iiijij

ii

i

k

kjikijij

ijij

i

k

kjikijij

ij

i

k

kjikijij

ji

ji

ji

jkkiijij

Provided

2.2 Derivation of Continuous and Discrete Hybrid Adams-Moulton Scheme

Case K=1

The matrix for this case is given below as

11

32

32

3210

3210

1

1

nn

nn

ununun

nnn

xx

xx

xxx

xxx

D and its equation (2.1) equivalence is

1100

nnunun fxfxhyxyxxy

By applying the set of formulae (2.6) - (2.9, we have

0

0

1

1

)10.2.....(......................................................................,.........4,3,2,1,

4141

3131

2121

1111

11

dl

dl

dl

dl

idl ii

Now using



25

3

11

1414

2

11

13

13

11

1212

4

3

2

1,4,3,2,1,

n

n

n

ijijij

xl

du

j

xl

du

j

xl

du

j

ijldu

Using

1,01

2,4,1.01

2,3

2,2

,

12414242

32

12313232

22

12212222

1

22222

1

1

n

n

nn

k

kik

j

k

kjikijij

xuldl

jixl

uldl

ji

uhxuhxl

uldl

uldl

ji

ijuldl

Using

,2,1,,1

1

ijiluldu ii

i

k

kjikijij



26

222

33

22

14212424

33

1.

4,2

2

1.2

22

13212323

4,1

huuhxx

uh

xuhx

l

uldu

ji

uhn

x

uh

nxuh

nx

l

uldu

ji

nn

n

Again using

hu

uhxhx

ululdl

ji

uhuhxx

ululdl

ji

jiuldl

nn

nn

i

k

kjikijij

2

222

3,4

22

3,3

233213414343

233213313333

1

1

,

Using

uhx

uh

huuhxxx

l

ululdu

jijiluldu

n

nnn

ii

i

k

kjikijij

3

333

,4,3,,

2222

33

2432143134

34

1

1



27

Using

222

2222

3443244214414444

1

1

,

2323

32333

1,4,

huuhh

uhxuhhhuuhxxhx

ulululdl

jiijuldl

nnnn

i

k

kjikijij

Using (2.7) with

0

4

0

3

0

2

1

1

11

141

14

11

131

13

11

121

12

11

111

11

1

l

ee

j

l

ee

j

l

ee

j

l

eee

ji

ij



28

44

1

3143

1

2142

1

1141411

41

22

33

1

2132

1

1131311

31

22

1121211

21

1

1

1

1,4

111

110

1,2

.4,3,2,1;4,3,2,1,

l

elelelee

ji

huuhuh

l

elelee

uhuh

l

elee

ji

jilelee ij

i

k

kjikijij

32

2

22

23

2

23

12

1

huu

hu

huuhh

uh

i = 2 = j

33

1

2232

1

1231321

32

22

1

1221221

22

2,3

1

l

elelee

ji

uh

l

elee



29

4,3

23

2

23

2

3,4

1

3

00.10

4,2

00.10

3,2

23

2

23

21

2,4

111

2

2

33

1

3343

1

2342

1

1341431

43

33

1

2332

1

1331331

33

22

1

1421241

24

22

22

1

1321231

23

22

2

22

44

1

3243

1

2242

1

1241421

42

22

ji

uhu

u

hu

uh

u

l

elelelee

ji

uh

l

elelee

ji

uh

l

elee

ji

l

l

elee

ji

huu

hu

hu

hu

uh

l

elelelee

ji

huuhuh



30

2

44

1

3443

1

2442

1

1441441

44

33

1

2432

1

1431341

34

23

1

4

00

hu

l

elelelee

ji

uh

l

elelee

C values are now computed thus;

2

1

4444

2

1

4343

22

1

4242

22

1

4141

23

1

4

23

2

3,4

23

2

2,4

23

2

huec

ji

uuh

uec

ji

uhuec

ji

uhuec

Using

4234

1

3232

32

2222

4134

1

3131

4

1

2,3

23

36

23

23

1

1,3

cuec

ji

uhu

hx

uhuuhx

hu

cuec

ji

cuec

n

n

k

kjikijij



31

2

4434

1

3434

2

2

2

4234

1

3333

32

3222

23

3

4,3

23

233

23

23

1

3

23

23

23

321

hu

uhx

cuec

ji

uuh

xuhu

uuh

uuhx

uh

cuec

ji

uhu

xh

uhu

uhx

hu

n

n

n

n

n

Now with

ji

uhu

xx

uhu

huuhxx

uhu

hxuhx

uh

cucuec

ji

cuec

nn

nnnn

k

k

kjikijij

2

23

6

23

233

23

3621

1,2

32

1

32

222

32

41242123

1

2121

1

3

1

24433323

1

2323

22

1

22

3232

42243223

1

2122

3,2

23

6

323

2

23

232

1

uccuec

ji

uhu

xx

huxxuhuhux

xhuhx

uh

cucuec

nn

unn

n

nn



32

uh

uhxx

huhuuhx

hu

uhxuhx

cucuec

ji

uuh

hxhuhxuxx

uuh

uhuuhxx

uuh

hxuxhuuhx

nn

nn

n

nnnn

nnnn

n

23

23

23

13

23

32

4,2

23

236

23

233

23

3632

2

2

22

2

44243423

1

2424

2

222

1

2

222

2

2

Now consider

uuh

ux

uuh

hxuxhux

uuh

uhhuhxuhxuxxx

cucucuec

ji

uhu

xhx

uhux

uhu

xx

uhu

xxx

cucucuec

ji

uhu

xhxhuhu

uhux

uhu

hxx

uhu

xxx

cucucuec

ji

cueC

nnnnnnnn

n

nn

n

nn

nnn

nn

nn

nnn

n

k

k

kjikijij

23

2

23

363

23

32636

3,1

23

23

23

2

23

6

23

6

2,1

23

2323

23

2

23

23

23

61

1

2

3

2

2

2

222222

431433132312

1

1313

32

32

32

3

32

2

1

2

32

1

421434132212

1

1212

32

323332

32

3

32

2

32

1

411431132112

1

1111

2

2

1



33

4,1

23

23232

32223222

ji

uuh

xhuxhxuuxhxuhx nnnnnn

uh

xx

uh

x

uh

uhxx

uh

huxxx

cucucuec

unn

nnnnnn

23

2323

3

23

23

2

2

2

3

2

2

2

2

441434132412

1

1414

The continuous scheme coefficients column by column for c values are now computed.

Substituting these values into the general form (2.10) yield the desired continuous schemes.

uuh

hxxuuxxhuhxxu

uuh

xu

uuh

hxuxhu

uuh

xuhhuhxuhxuhxuxxx

uuh

xhuxhxuhxuhxuhx

xcxcxccx

uhu

xxhxx

uhu

x

uhu

xhx

uhu

xxhx

hu

xhx

xcxcxccx

uhu

uhuxxhxx

uhu

x

uhu

xhx

uhu

xhxx

uhu

xhxhuhu

xcxcxccx

nnn

nn

nnnnnn

nnnnnn

n

n

nnnnn

nn

nnnnn

23

3232

23

2

23

363

23

323636

23

232323

23

32

23

2

23

36

23

6623

23

2332

23

2

23

36

23

66

23

2323

2

2223

2

3

2

22

2

2222223

322222222

3

34

2

3323130

32

2

3

32

2

32

2

32

2

32

32

3

24

2

2322121

32

3223

32

3

32

2

32

2

32

323322

3

41

2

3121110



34

uh

xxuhxx

uh

x

uh

xuhx

uh

xhuxx

uh

xhux

xcxcxccx

nn

nnnnn

23

2323

3

23

23

23

2

23

2

3

2

2

2

2

2

32

3

44

2

3424141

Hence forth,

11.2....2323

3232

23

32

23

2332

12

23

2

22223

.

32

23

32

3223

nnn

nnnn

nnn

nn

fuh

xxuhxxf

uuh

xxhuuxxhuxxu

unuhu

xxhxxy

uhu

uhuxxhxxxy y

Where (2.11) is the continuous form of one step Adams-Moulton scheme for k=1. If

(2.11) is evaluated at 1nx and a substitution 2

1u is made, the result is

hencefh

fh

yyxy nnnnn ,1144

22

1

12.2........................................................................................................4

2 11 21 nnnnn ff

hyyy

If on the other hand, equation (2.11) is differentiated with respect to x and then evaluated at

21

n

xx, the result obtained is

bfffh

yy

ffyh

yh

x

nnnn

n

nnnnny

12.2...........................................................................8524

8

1

8

533

21

21

21

1

2

1

,1

1

The schemes (2.12a) and (2.12b) can be referred to as (2.12) and are indeed of interpolation

polynomial of case k =1

We consider another system of matrix of the same case k=1 where the schemes shall be derived

at the interpolation points



35

2

2

11

2

32

3210

3210

3210

1

unun

nn

nn

nnn

xx

xx

xx

xxx

D

The general forms of this system of matrix is given as :

13.2............................................................1100 unnnnn fxfxfxhyxxy

As seen in the first case using the same set of formulae (2.6) – (2.9), we have the values of c given as

follows:

1

1

1

0

12

2

12

2

2

2

0

13

1

13

13

1

;0

11

2

1

24

2

2

23

2

22

22

21

234

233

232

31

244

243

242

41

21

c

uuh

xxc

uh

xhuxc

uh

hxhuxxuhc

c

uuh

xc

uh

xuhc

uh

huhxc

c

uuhc

uuh

uc

uuh

uc

c

nn

nn

nnn

n

n

n



36

16

23

6

2336

2

32

13

2

3222

12

xuh

xhxc

uh

xhxhuxhuxc

nn

nnnn

Following the same procedure as in the first case, the coefficients for the continuous schemes,

xandx jj are obtained and by similar substitution into equation (2.13), the desired continuous scheme

is obtained as follows;

14.2..............................16

32

16

32

16

1613121

2

23

12

223

2

2223

unnn

nnn

nnnn

n

fuuh

xxhxxf

uuh

xxhuxxu

fuuh

xxuuhxxuhxxuyxy

Putting u = ½ and proceeding to get our discrete forms as did earlier by evaluating (2.14) at these two

different points 1 nxx and 1 nn xx , we have:

a. at x = xn = xn+1, the result is

afff

hyy

hffh

fh

yy

nnnnn

nnnnn

15.2...........................................................................46

6

4

66

21

21

11

11

b. at x = x + xn+ ½ , the result is

bfff

hyy

nnnnn15.2......................................................................85

24 21

21 1

The collocation schemes of D yields just the same equation in (2.15b) . These are the hybrid collocation

schemes of the case k=1

Equation (2.15a) and (2.15b) can be referred to as (2.15)

We also considered two different systems of matrices D3 and D4 on the derivation of both hybrid

interpolation and the collocation where k = 2



37

32

3

2

2

22

3

1

2

11

32

4

1

3

1

2

11

3

43210

43210

43210

43210

1

ununun

nnn

nnn

nnn

nnnn

xxx

xxx

xxx

xxx

xxxx

D

The general forms of D3 is given as

unnnnnn fxfxfxfxhyxxy

2211011

Appling the set of formulae (2.6)-(2.9) on this system gives the c-entries as follows

26

3

112

2412

24

12124

124

1

8

1

1,0

344

343

342

353

352

11314151

uh

huhxc

uh

huxc

uh

hxuhc

uuuhc

uhc

cccc

n

n

n

3

22233

22

3

22

35

3

22

34

3

22

33

3

222

32

3

1

45

2

2332

122

263

24

4323

12

4322

4

22622

12

uh

xhuxhuxhxxuhc

uuuh

hhxxc

uh

hxxhxxc

uh

hxxhuxuhc

uh

huxhhxuhuhc

uuuh

xc

nnnn

nn

nnnn

nnn

nn

n



38

124

44

224

23644

112

8412853

24

3101212341824

1

23

1

22

3

44223

15

3

4442233

14

3

332244

13

3

44222343223

12

3

322

25

3

322

23

uuuh

hxhxhxc

uh

huhxhuxhuxhxc

hu

hxhuxhuxuxhhxc

uh

huxhhxhxxhuxhuxhuxc

uuh

xhxhxc

uh

xhxhuxhuxc

nnn

nnnn

nnnn

nnnnnn

nnn

nnnn

As in the previous cases, continuous schemes coefficients sxi '

and the sxi ' are obtained

and

substituted into the general form (2.16) to obtain the desired continuous schemes as follows;

17.2.............................................................................214

44

224

126143

112

8512243

24

10324121812431

3

42234

3

2

42234

3

1

42234

3

432234

1

uuuh

fhhxxhxxxx

uh

fhuxxuhxxuhxx

uh

fhuxxuhxxhuxx

uh

fhuxxuhhxxuhxxuxxyxy

unnnn

nnnn

nnnn

nnnnnn

Evaluating (2.17) at 2 nxx 2

3u , the resulting scheme is given as:

afffh

yynnnnn 18.2....................................................................................................4

6 232112

If (2.17) is evaluated at points unxx and 2

3u , we have

bffffh

yynnnnnn

18.2.............................................................................56546192 2

32

3 211



39

However, if the point 2

3u is maintained and (2.17) is evaluated at x = xn , the result is another discrete

scheme of the form:

cffffh

yynnnnnn 18.2.........................................................................................472

6 23211

Equations (2.18a), (2.18b) and (2.18c) are referred to as (2.18)

Taking on another system of matrix D4 of an interpolation scheme, where k = 2:

3

2

2

22

3

1

2

11

32

432

4

1

3

1

2

!1

4

43210

43210

43210

1

1

nnn

nnn

nnn

unununun

nnnn

xxx

xxx

xxx

xxxx

xxxx

D

the general form of the matrix is given as:

19.2.......................................221011

nnnnunun fxfxfxhyxyxxy

As in the previous cases, matrix C is determined with the columns simplification yielding the continuous

schemes:

20.2..............1212

13112212

1213

222312122

12252121

2

121

1212

32

2

22

1

3

1

24

1

32

1

32

1

24

1

2

324

1422

1

22

1

4

1

422

42222

1

4

1

huu

fhxxuuhxxuuxxu

huuu

fhuuhxxuuuxxuu

fhuuxxuhuuu

yhxxxx

huuu

yhuuuhxxxxxy

nnnn

nnn

nnnnn

unnn



40

Desired hybrid point of elevation remains 2

3U

and so (2.20) is evaluated to yield

afffh

yyy nnnnnn 21.2................................................................193212

1697 21122

3

at the point 2 nxx

If consideration at the point nxx

is made, we have:

bfffh

yyy nnnnnn 21.2........................................................................34894

1679 3112

3

Finding the differential expression with respect to x for (2.20) and evaluating the differential coefficient at

1 nxx , the result becomes:

cffffh

yynnnnnn 21.2..........................................................................56546

192 23

23 211

As usual, equation (2.21a),(2.21b) and (2.21c) are referred to as (2.21)

3.0 Convergence Analysis

This section shows the validity and consistency of the derived scheme in section two . The tools for the

assignment would be the familiar investigation of the zero stability by finding the order and error constant

of the each of the schemes.

3.1 Definitions

( a) The scheme (1.1) is said to be zero stable if no root of the polynomial

k

i

i

idp0

has modulo

greater than one and every root with modulo one must be distinct or simple.

( b) The order p and error constant 1pc for (1.1) could be defined thus:

If 0,0... 110 pp cccc ,then the principal local term error at knx is .11

1 n

pp

p xyhc



41

1.3.........................21

!1

1

...21!

1

11

2

1

1

1

0

1

210

v

q

k

qqqq

qqqq

q

tvtktttq

tktttq

c

(c) A numerical method (1.1) is said to be consistent if ,1p where p is the order of the method.

3.1 Example

We take on the derived schemes in section 2 above, showing that each of the schemes is in conformity to

the definition above or otherwise.

Example 3.1.1

One of the derived schemes for step number k=1 is given as:

affh

yyy nnnnn 12.2.........................................................................................4

2 11 21

From equation (2.12a)

4

1,

4

1,2,1,1 1010 2

1



42

192

1

4

1.

6

12

6

11

24

1

!3

1

2

11

!4

1

08

12

8

11

6

1

!2

1

2

1

!3

1

04

12

2

11

2

1

2

1

!2

1

0114

1

2

121

2

1

0211

1

4

4

1

3

13

1

2

12

1011

100

21

21

21

21

21

c

c

c

c

c

Therefore the order p = 3 and

1.

01

021

21

21

21

2101

ei

p

Hence, the schemes is zero stable.

Example 3.1.2

Another scheme of k = 1 was given as:

2

12

1 8524

1

nnnnn fffh

yy

Here, 31

241

1245

002

12

1 ,,,,1,1 and and so



43

384

1

144

1

144

1

384

1

2

1

!3

1

2

1!

4

1

048

1

48

1

24

1

2

1

48

1

12

1

24

1

2

1

48

1)

2

1(

2

1

2

1!

3

1

0)2

1(

3

1

24

1

8

1

2

1

2

1!

2

1

;024

12

2

1

2

1

;011

2

1

3

1

2

1

4

4

2

1

2

1

2

1

3

3

2

11

2

2

101

0

21

21

21

c

c

c

c

c

So that the order p = 3 and ;384

11 pc we seek the root of the characteristics polynomial as follows:

1

01 21

210

p

Hence, by definitions 3.1, we concluded that the method is zero stable.

We consider the scheme 2

32

3 56546192

211

nnnnnnffff

hyy

Here, 24

7,,

96

23,

192

5,

192

1,1,1

23

23 1201 and

Then observe that 0.................................................................................. 410 ccc

However,

3

4

23

2

4

1

5

23

15

1087.4

307224

3692

32

211

120

1

3072

4236

192

80

192

46

24

1

32

2431

120

1

2!4

1

!5

12

32

3

c

The order p=4 and .1087.4 3

1

pc the absolute root for the polynomial is thus computed;



44

1,,,1,,0

01.

0

21

23

23

23 1

soandor

ei

p

Hence, the method is zero stable.

4.0 Conclusion

There is no doubt that the schemes are consistent and zero stable and could also be used by Numerical

Analysts to solve differential equations experimentally. The obtained result in comparism with the

theoretical result would also be of great importance in future research work.

References

Alagbe, S.A. (1999), “Derivation of a Class Multiplier Step Methods”, B.Sc. Project , Department of

Mathematics, University of Jos,

Awe, K.D. (1997), “Application of Adams Moulton Methods and its application in Block Forms”,

B.Sc Project, Department of Mathematics, University of Jos

Ciarlet, P.G. & Lions, J.L (1989). “Finite Element Method` Handbook of Numerical Analysis”. Vol.11

Fatunla S. O. (1988), “Computer Science and Scientific Computational Method for IVP in ODE”.

Academic Press Inc.

Gragg,W.B & Stetter H.J (1964), “Generalized Multistep Predictor-Corrector Methods”.

J-Assoc.Comp.Vol11, pp188-200,MR 28 Comp 4680

James, R. e tal (1995), “The Mathematics of Numerical Analysis”. Lecture in

Applied Mathematics.Vol XXXII.

Kopal, Z.(1956), “Numerial Analysis”. Chapman and Hall, London.

Lapidus L.B & Schiesser W.E. (1976), “Numerical Methods for Differential System”.

Academic Press Inc.

Onumanyi, P. et al (1994), “New Linear Multistep Methods for First Order Ordinary Initial Value

Problems“. Jour. Nig. Maths Society, 13 pp37 – 52

Sirisena, U.W. (1997), “A Reformation of the Continous General Linear Multistep Method by Matrix

Inversion for First Order Initial Value Problems”. Ph.D Thesis, Department of Mathematics, University of

Ilorin, Ilorin (Unpublished)

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