In charlotte-ngs/ZLHS2016:
knitr::opts_chunk$set(echo = FALSE, results = 'asis')
devtools::load_all()
Inverse von $A$
- LDL-Zerlegung von $A$ führt zu
$$A^{-1} = L^{-1} * D^{-1} * (L^T)^{-1}$$
- Matrix $D$ abhängig von Inzuchtkoeffizienten
- Inzuchtkoeffizienten sind auf Diagonalen von $A$
- Aufstellen der ganzen $A$???
Berechnung der Inzuchtkoeffizienten
- Cholesky-Zerlegung von $A$
- Pfadkoeffizienten aufgrund der Definition, alle Pfade über gemeinsame Ahnen (nicht weiter ausgeführt)
Cholesky-Zerlegung von $A$
- Definition: Zerlegung von $A$ in
$$A = R * R^T$$
wobei $R$ linke untere Dreiecksmatrix
- In R: Funktion
chol() liefert transponierte von $R$
Kleines Beispiel
nAnzAni <- 3
matA <- matGetMatElem(psBaseElement = "a", pnNrRow = nAnzAni, pnNrCol = nAnzAni)
matR <- matLowerTri(psBaseElement = "r", pnNrRow = nAnzAni, pnNrCol = nAnzAni)
cat("$$\\left[")
cat(paste(sGetTexMatrix(pmatAMatrix = matA), collapse = "\n"))
cat("\\right] \n")
cat(" = \\left[")
cat(paste(sGetTexMatrix(pmatAMatrix = matR), collapse = "\n"))
cat("\\right] \n")
cat(" * \\left[")
cat(paste(sGetTexMatrix(pmatAMatrix = t(matR)), collapse = "\n"))
cat("\\right] \n")
cat("$$\n")
- Diagonalelemente
\begin{eqnarray}
a_{11} &=& r_{11}^2 \nonumber\
a_{22} &=& r_{21}^2 + r_{22}^2\nonumber\
a_{33} &=& r_{31}^2 + r_{32}^2 + r_{33}^2\nonumber
\label{MatADiag}
\end{eqnarray}
Rekursive Berechnung von $R$
- Gleichsetzen der LDL- und der Cholesky-Zerlegung
$$A = R * R^T = L * D * L^T$$
- Sei $R = L * S$ und setzen das ein, dann folgt
$$A = L * D * L^T = L * S * (L * S)^T = L * S * S^T * L^T$$
- Somit ist $D = S * S^T$
wobei $S$ eine Diagonalmatrix mit $s_{ii} = \sqrt{d_{ii}}$
Kleines Beispiel
$$
matL <- matLowerTri(psBaseElement = "l", pnNrRow = nAnzAni, pnNrCol = nAnzAni, pvecDiag = 1)
matS <- matDiag(psBaseElement = "s", pnNrRow = nAnzAni, pnNrCol = nAnzAni)
cat("\\left[")
cat(paste(sGetTexMatrix(pmatAMatrix = matR), collapse = "\n"))
cat("\\right] = \n")
cat("\\left[")
cat(paste(sGetTexMatrix(pmatAMatrix = matL), collapse = "\n"))
cat("\\right] * \n")
cat("\\left[")
cat(paste(sGetTexMatrix(pmatAMatrix = matS), collapse = "\n"))
cat("\\right]\n")
$$
-
Diagonalelemente: $$r_{ii} = s_{ii} = \sqrt{d_{ii}}$$
-
Off-Diagonal: $$r_{ij} = l_{ij} * s_{jj} = {1\over 2}\ (r_{sj} + r_{dj})$$
wobei $l_{ij} = {1\over 2}\ (l_{sj} + l_{dj})$
Unser Beipielpedigree
suppressPackageStartupMessages(library(pedigreemm))
nNrAni <- 6
pedEx1 <- pedigree(sire = c(NA,NA,1,1,4,5),
dam = c(NA,NA,2,NA,3,2),
label = as.character(1:nNrAni))
print(pedEx1)
- Aufgabe: Berechne Diagonalelemente von $A$
Tier 1
- Element $a_{11}$
- keine bekannten Eltern
- somit nicht ingezüchtet $\rightarrow F_1 = 0$ und
matA <- as.matrix(getA(pedEx1))
vecA <- as.vector(diag(matA))
matR <- t(chol(matA))
$$a_{11} = 1 + F_1 = r vecA[1]$$
- Da $a_{11} = r_{11}^2$ folgt $r_{11} = 1$
Tier 2
- Anwendung der Formel für Diagonalelemente $a_{ii}$
$$a_{22} = r_{21}^2 + r_{22}^2$$
- Eltern von Tier $2$ unbekannt, $\rightarrow r_{sj} = 0$ und $r_{dj} = 0$
$$r_{21} = r matR[2,1]$$.
- Eltern von Tier $2$ beide unbekannt
$$r_{22} = r matR[2,2]$$
$$a_{22} = r_{21}^2 + r_{22}^2 = r matR[2,1]^2 + r matR[2,2]^2 = r matR[2,1]^2 + matR[2,2]^2$$
Tier 3
- Diagonalelement $a_{33}$ aus Elementen $r_{31}$, $r_{32}$ und $r_{33}$
- Tier $3$ hat Eltern $1$ und $2$
$$r_{31} = {1\over 2}(r_{11} + r_{21}) = {1\over 2}(r matR[1,1] + r matR[2,1]) = r 0.5*(matR[1,1] + matR[2,1])$$
stopifnot(all.equal(0.5*(matR[1,1] + matR[2,1]), matR[3,1]))
$$r_{32} = {1\over 2}(r_{12} + r_{22}) = {1\over 2}(r matR[1,2] + r matR[2,2]) = r 0.5*(matR[1,2] + matR[2,2])$$
stopifnot(all.equal(0.5*(matR[1,2] + matR[2,2]), matR[3,2]))
$$r_{33} = \sqrt{1 - 0.25(a_{11} + a_{22})} = \sqrt{1 - 0.25(r vecA[1] + r vecA[2])}
= \sqrt{r 1-0.25*(vecA[1]+vecA[2])}
= r sqrt(1-0.25*(vecA[1]+vecA[2]))$$
stopifnot(all.equal(sqrt(1-0.25*(vecA[1]+vecA[2])), matR[3,3]))
$$a_{33} = r_{31}^2 + r_{32}^2 + r_{33}^2
= r matR[3,1]^2 + r matR[3,2]^2 + r matR[3,3]^2
= r matR[3,1]^2 + matR[3,2]^2 + matR[3,3]^2
$$
stopifnot(all.equal(matR[3,1]^2 + matR[3,2]^2 + matR[3,3]^2, vecA[3]))
Tier 4
- Mutter unbekannt
- Inzuchtkoeffizient $F_4 = 0$ und $a_{44} = 1$
- da Tier $4$ Nachkommen hat, müssen Elemente auf Zeile $4$ in Matrix $R$ berechnet werden
$$r_{41} = {1\over 2} * r_{11} = {1\over 2} * r matR[1,1] = r 0.5 * matR[1,1]$$
stopifnot(all.equal(0.5 * matR[1,1], matR[4,1]))
$$r_{42} = {1\over 2} * r_{12} = {1\over 2} * r matR[1,2] = r 0.5 * matR[1,2]$$
stopifnot(all.equal(0.5 * matR[1,2], matR[4,2]))
$$r_{43} = {1\over 2} * r_{13} = {1\over 2} * r matR[1,3] = r 0.5 * matR[1,3]$$
stopifnot(all.equal(0.5 * matR[1,3], matR[4,3]))
$$r_{44} = \sqrt{1 - {1\over 4} * a_{11}}
= \sqrt{1 - {1\over 4} * r vecA[1]}
= r sqrt(1 - 0.25*vecA[1])
$$
stopifnot(all.equal(sqrt(1 - 0.25*vecA[1]), matR[4,4]))
Tier 5
$$r_{51} = {1\over 2}(r_{41} + r_{31}) = {1\over 2}(r matR[4,1] + r matR[3,1]) = r 0.5*(matR[4,1] + matR[3,1])$$
stopifnot(all.equal(0.5*(matR[4,1] + matR[3,1]), matR[5,1]))
$$r_{52} = {1\over 2}(r_{42} + r_{32}) = {1\over 2}(r matR[4,2] + r matR[3,2]) = r 0.5*(matR[4,2] + matR[3,2])$$
stopifnot(all.equal(0.5*(matR[4,2] + matR[3,2]), matR[5,2]))
$$r_{53} = {1\over 2}(r_{43} + r_{33}) = {1\over 2}(r matR[4,3] + r matR[3,3]) = r 0.5*(matR[4,3] + matR[3,3])$$
stopifnot(all.equal(0.5*(matR[4,3] + matR[3,3]), matR[5,3]))
$$r_{54} = {1\over 2}(r_{44} + r_{34}) = {1\over 2}(r matR[4,4] + r matR[3,4]) = r 0.5*(matR[4,4] + matR[3,4])$$
stopifnot(all.equal(0.5*(matR[4,4] + matR[3,4]), matR[5,4]))
$$r_{55} = \sqrt{1 - 0.25(a_{44}+a_{33})} = \sqrt{1 - 0.25(r vecA[4] + r vecA[3])}
= r sqrt(1 - 0.25*(vecA[4] + vecA[3]))
$$
stopifnot(all.equal(sqrt(1 - 0.25*(vecA[4] + vecA[3])), matR[5,5]))
$$a_{55} = r_{51}^2 + r_{52}^2 + r_{53}^2 + r_{54}^2 + r_{55}^2
= r matR[5,1]^2 + matR[5,2]^2 + matR[5,3]^2 + matR[5,4]^2 + matR[5,5]^2
$$
stopifnot(all.equal(matR[5,1]^2 + matR[5,2]^2 + matR[5,3]^2 + matR[5,4]^2 + matR[5,5]^2, vecA[5]))
Tier 6
$$r_{61} = {1\over 2}(r_{51} + r_{21}) = {1\over 2}(r matR[5,1] + r matR[2,1]) = r 0.5*(matR[5,1] + matR[2,1])$$
stopifnot(all.equal(0.5*(matR[5,1] + matR[2,1]), matR[6,1]))
$$r_{62} = {1\over 2}(r_{52} + r_{22}) = {1\over 2}(r matR[5,2] + r matR[2,2]) = r 0.5*(matR[5,2] + matR[2,2])$$
stopifnot(all.equal(0.5*(matR[5,2] + matR[2,2]), matR[6,2]))
$$r_{63} = {1\over 2}(r_{53} + r_{23}) = {1\over 2}(r matR[5,3] + r matR[2,3]) = r 0.5*(matR[5,3] + matR[2,3])$$
stopifnot(all.equal(0.5*(matR[5,3] + matR[2,3]), matR[6,3]))
$$r_{64} = {1\over 2}(r_{54} + r_{24}) = {1\over 2}(r matR[5,4] + r matR[2,4]) = r 0.5*(matR[5,4] + matR[2,4])$$
stopifnot(all.equal(0.5*(matR[5,4] + matR[2,4]), matR[6,4]))
$$r_{65} = {1\over 2}(r_{55} + r_{25}) = {1\over 2}(r matR[5,5] + r matR[2,5]) = r 0.5*(matR[5,5] + matR[2,5])$$
stopifnot(all.equal(0.5*(matR[5,5] + matR[2,5]), matR[6,5]))
$$r_{66} = \sqrt{1 - 0.25(a_{55}+a_{22})}
= \sqrt{1 - 0.25(r vecA[5]+ r vecA[2])}
= r sqrt(1 - 0.25*(vecA[5]+ vecA[2]))
$$
stopifnot(all.equal(sqrt(1 - 0.25*(vecA[5]+ vecA[2])), matR[6,6]))
$$a_{66} = r_{61}^2 + r_{62}^2 + r_{63}^2 + r_{64}^2 + r_{65}^2 + r_{66}^2
= r matR[6,1]^2 + matR[6,2]^2 + matR[6,3]^2 + matR[6,4]^2 + matR[6,5]^2 + matR[6,6]^2
$$
stopifnot(all.equal(matR[6,1]^2 + matR[6,2]^2 + matR[6,3]^2 + matR[6,4]^2 + matR[6,5]^2 + matR[6,6]^2, vecA[6]))
Zusammenfassung der Ergebnisse
dfSummaryResult <- data.frame(Tier = 1:nNrAni,
Diagonalelement = vecA,
Inzuchtkoeffzient = vecA-1,
stringsAsFactors = FALSE )
knitr::kable(dfSummaryResult)
charlotte-ngs/ZLHS2016 documentation built on May 13, 2019, 3:33 p.m.
R Package Documentation
Browse R Packages
We want your feedback!
Note that we can't provide technical support on individual packages. You should contact the package authors for that.
knitr::opts_chunk$set(echo = FALSE, results = 'asis') devtools::load_all()
Inverse von $A$
- LDL-Zerlegung von $A$ führt zu
$$A^{-1} = L^{-1} * D^{-1} * (L^T)^{-1}$$
- Matrix $D$ abhängig von Inzuchtkoeffizienten
- Inzuchtkoeffizienten sind auf Diagonalen von $A$
- Aufstellen der ganzen $A$???
Berechnung der Inzuchtkoeffizienten
- Cholesky-Zerlegung von $A$
- Pfadkoeffizienten aufgrund der Definition, alle Pfade über gemeinsame Ahnen (nicht weiter ausgeführt)
Cholesky-Zerlegung von $A$
- Definition: Zerlegung von $A$ in
$$A = R * R^T$$
wobei $R$ linke untere Dreiecksmatrix
- In R: Funktion
chol()liefert transponierte von $R$
Kleines Beispiel
nAnzAni <- 3 matA <- matGetMatElem(psBaseElement = "a", pnNrRow = nAnzAni, pnNrCol = nAnzAni) matR <- matLowerTri(psBaseElement = "r", pnNrRow = nAnzAni, pnNrCol = nAnzAni) cat("$$\\left[") cat(paste(sGetTexMatrix(pmatAMatrix = matA), collapse = "\n")) cat("\\right] \n") cat(" = \\left[") cat(paste(sGetTexMatrix(pmatAMatrix = matR), collapse = "\n")) cat("\\right] \n") cat(" * \\left[") cat(paste(sGetTexMatrix(pmatAMatrix = t(matR)), collapse = "\n")) cat("\\right] \n") cat("$$\n")
- Diagonalelemente \begin{eqnarray} a_{11} &=& r_{11}^2 \nonumber\ a_{22} &=& r_{21}^2 + r_{22}^2\nonumber\ a_{33} &=& r_{31}^2 + r_{32}^2 + r_{33}^2\nonumber \label{MatADiag} \end{eqnarray}
Rekursive Berechnung von $R$
- Gleichsetzen der LDL- und der Cholesky-Zerlegung
$$A = R * R^T = L * D * L^T$$
- Sei $R = L * S$ und setzen das ein, dann folgt
$$A = L * D * L^T = L * S * (L * S)^T = L * S * S^T * L^T$$
- Somit ist $D = S * S^T$
wobei $S$ eine Diagonalmatrix mit $s_{ii} = \sqrt{d_{ii}}$
Kleines Beispiel
$$
matL <- matLowerTri(psBaseElement = "l", pnNrRow = nAnzAni, pnNrCol = nAnzAni, pvecDiag = 1) matS <- matDiag(psBaseElement = "s", pnNrRow = nAnzAni, pnNrCol = nAnzAni) cat("\\left[") cat(paste(sGetTexMatrix(pmatAMatrix = matR), collapse = "\n")) cat("\\right] = \n") cat("\\left[") cat(paste(sGetTexMatrix(pmatAMatrix = matL), collapse = "\n")) cat("\\right] * \n") cat("\\left[") cat(paste(sGetTexMatrix(pmatAMatrix = matS), collapse = "\n")) cat("\\right]\n")
$$
-
Diagonalelemente: $$r_{ii} = s_{ii} = \sqrt{d_{ii}}$$
-
Off-Diagonal: $$r_{ij} = l_{ij} * s_{jj} = {1\over 2}\ (r_{sj} + r_{dj})$$
wobei $l_{ij} = {1\over 2}\ (l_{sj} + l_{dj})$
Unser Beipielpedigree
suppressPackageStartupMessages(library(pedigreemm)) nNrAni <- 6 pedEx1 <- pedigree(sire = c(NA,NA,1,1,4,5), dam = c(NA,NA,2,NA,3,2), label = as.character(1:nNrAni)) print(pedEx1)
- Aufgabe: Berechne Diagonalelemente von $A$
Tier 1
- Element $a_{11}$
- keine bekannten Eltern
- somit nicht ingezüchtet $\rightarrow F_1 = 0$ und
matA <- as.matrix(getA(pedEx1)) vecA <- as.vector(diag(matA)) matR <- t(chol(matA))
$$a_{11} = 1 + F_1 = r vecA[1]$$
- Da $a_{11} = r_{11}^2$ folgt $r_{11} = 1$
Tier 2
- Anwendung der Formel für Diagonalelemente $a_{ii}$
$$a_{22} = r_{21}^2 + r_{22}^2$$
- Eltern von Tier $2$ unbekannt, $\rightarrow r_{sj} = 0$ und $r_{dj} = 0$
$$r_{21} = r matR[2,1]$$.
- Eltern von Tier $2$ beide unbekannt
$$r_{22} = r matR[2,2]$$
$$a_{22} = r_{21}^2 + r_{22}^2 = r matR[2,1]^2 + r matR[2,2]^2 = r matR[2,1]^2 + matR[2,2]^2$$
Tier 3
- Diagonalelement $a_{33}$ aus Elementen $r_{31}$, $r_{32}$ und $r_{33}$
- Tier $3$ hat Eltern $1$ und $2$
$$r_{31} = {1\over 2}(r_{11} + r_{21}) = {1\over 2}(r matR[1,1] + r matR[2,1]) = r 0.5*(matR[1,1] + matR[2,1])$$
stopifnot(all.equal(0.5*(matR[1,1] + matR[2,1]), matR[3,1]))
$$r_{32} = {1\over 2}(r_{12} + r_{22}) = {1\over 2}(r matR[1,2] + r matR[2,2]) = r 0.5*(matR[1,2] + matR[2,2])$$
stopifnot(all.equal(0.5*(matR[1,2] + matR[2,2]), matR[3,2]))
$$r_{33} = \sqrt{1 - 0.25(a_{11} + a_{22})} = \sqrt{1 - 0.25(r vecA[1] + r vecA[2])}
= \sqrt{r 1-0.25*(vecA[1]+vecA[2])}
= r sqrt(1-0.25*(vecA[1]+vecA[2]))$$
stopifnot(all.equal(sqrt(1-0.25*(vecA[1]+vecA[2])), matR[3,3]))
$$a_{33} = r_{31}^2 + r_{32}^2 + r_{33}^2
= r matR[3,1]^2 + r matR[3,2]^2 + r matR[3,3]^2
= r matR[3,1]^2 + matR[3,2]^2 + matR[3,3]^2
$$
stopifnot(all.equal(matR[3,1]^2 + matR[3,2]^2 + matR[3,3]^2, vecA[3]))
Tier 4
- Mutter unbekannt
- Inzuchtkoeffizient $F_4 = 0$ und $a_{44} = 1$
- da Tier $4$ Nachkommen hat, müssen Elemente auf Zeile $4$ in Matrix $R$ berechnet werden
$$r_{41} = {1\over 2} * r_{11} = {1\over 2} * r matR[1,1] = r 0.5 * matR[1,1]$$
stopifnot(all.equal(0.5 * matR[1,1], matR[4,1]))
$$r_{42} = {1\over 2} * r_{12} = {1\over 2} * r matR[1,2] = r 0.5 * matR[1,2]$$
stopifnot(all.equal(0.5 * matR[1,2], matR[4,2]))
$$r_{43} = {1\over 2} * r_{13} = {1\over 2} * r matR[1,3] = r 0.5 * matR[1,3]$$
stopifnot(all.equal(0.5 * matR[1,3], matR[4,3]))
$$r_{44} = \sqrt{1 - {1\over 4} * a_{11}}
= \sqrt{1 - {1\over 4} * r vecA[1]}
= r sqrt(1 - 0.25*vecA[1])
$$
stopifnot(all.equal(sqrt(1 - 0.25*vecA[1]), matR[4,4]))
Tier 5
$$r_{51} = {1\over 2}(r_{41} + r_{31}) = {1\over 2}(r matR[4,1] + r matR[3,1]) = r 0.5*(matR[4,1] + matR[3,1])$$
stopifnot(all.equal(0.5*(matR[4,1] + matR[3,1]), matR[5,1]))
$$r_{52} = {1\over 2}(r_{42} + r_{32}) = {1\over 2}(r matR[4,2] + r matR[3,2]) = r 0.5*(matR[4,2] + matR[3,2])$$
stopifnot(all.equal(0.5*(matR[4,2] + matR[3,2]), matR[5,2]))
$$r_{53} = {1\over 2}(r_{43} + r_{33}) = {1\over 2}(r matR[4,3] + r matR[3,3]) = r 0.5*(matR[4,3] + matR[3,3])$$
stopifnot(all.equal(0.5*(matR[4,3] + matR[3,3]), matR[5,3]))
$$r_{54} = {1\over 2}(r_{44} + r_{34}) = {1\over 2}(r matR[4,4] + r matR[3,4]) = r 0.5*(matR[4,4] + matR[3,4])$$
stopifnot(all.equal(0.5*(matR[4,4] + matR[3,4]), matR[5,4]))
$$r_{55} = \sqrt{1 - 0.25(a_{44}+a_{33})} = \sqrt{1 - 0.25(r vecA[4] + r vecA[3])}
= r sqrt(1 - 0.25*(vecA[4] + vecA[3]))
$$
stopifnot(all.equal(sqrt(1 - 0.25*(vecA[4] + vecA[3])), matR[5,5]))
$$a_{55} = r_{51}^2 + r_{52}^2 + r_{53}^2 + r_{54}^2 + r_{55}^2
= r matR[5,1]^2 + matR[5,2]^2 + matR[5,3]^2 + matR[5,4]^2 + matR[5,5]^2
$$
stopifnot(all.equal(matR[5,1]^2 + matR[5,2]^2 + matR[5,3]^2 + matR[5,4]^2 + matR[5,5]^2, vecA[5]))
Tier 6
$$r_{61} = {1\over 2}(r_{51} + r_{21}) = {1\over 2}(r matR[5,1] + r matR[2,1]) = r 0.5*(matR[5,1] + matR[2,1])$$
stopifnot(all.equal(0.5*(matR[5,1] + matR[2,1]), matR[6,1]))
$$r_{62} = {1\over 2}(r_{52} + r_{22}) = {1\over 2}(r matR[5,2] + r matR[2,2]) = r 0.5*(matR[5,2] + matR[2,2])$$
stopifnot(all.equal(0.5*(matR[5,2] + matR[2,2]), matR[6,2]))
$$r_{63} = {1\over 2}(r_{53} + r_{23}) = {1\over 2}(r matR[5,3] + r matR[2,3]) = r 0.5*(matR[5,3] + matR[2,3])$$
stopifnot(all.equal(0.5*(matR[5,3] + matR[2,3]), matR[6,3]))
$$r_{64} = {1\over 2}(r_{54} + r_{24}) = {1\over 2}(r matR[5,4] + r matR[2,4]) = r 0.5*(matR[5,4] + matR[2,4])$$
stopifnot(all.equal(0.5*(matR[5,4] + matR[2,4]), matR[6,4]))
$$r_{65} = {1\over 2}(r_{55} + r_{25}) = {1\over 2}(r matR[5,5] + r matR[2,5]) = r 0.5*(matR[5,5] + matR[2,5])$$
stopifnot(all.equal(0.5*(matR[5,5] + matR[2,5]), matR[6,5]))
$$r_{66} = \sqrt{1 - 0.25(a_{55}+a_{22})}
= \sqrt{1 - 0.25(r vecA[5]+ r vecA[2])}
= r sqrt(1 - 0.25*(vecA[5]+ vecA[2]))
$$
stopifnot(all.equal(sqrt(1 - 0.25*(vecA[5]+ vecA[2])), matR[6,6]))
$$a_{66} = r_{61}^2 + r_{62}^2 + r_{63}^2 + r_{64}^2 + r_{65}^2 + r_{66}^2
= r matR[6,1]^2 + matR[6,2]^2 + matR[6,3]^2 + matR[6,4]^2 + matR[6,5]^2 + matR[6,6]^2
$$
stopifnot(all.equal(matR[6,1]^2 + matR[6,2]^2 + matR[6,3]^2 + matR[6,4]^2 + matR[6,5]^2 + matR[6,6]^2, vecA[6]))
Zusammenfassung der Ergebnisse
dfSummaryResult <- data.frame(Tier = 1:nNrAni, Diagonalelement = vecA, Inzuchtkoeffzient = vecA-1, stringsAsFactors = FALSE ) knitr::kable(dfSummaryResult)
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