Problem:
Score : 200 points

Problem Statement

We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}.

The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet.

Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal.

Constraints

All values in input are integers.

1 \leq A_{i, j} \leq 100

A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2))

1 \leq N \leq 10

1 \leq b_i \leq 100

b_i \neq b_j (i \neq j)

Input

Input is given from Standard Input in the following format:

A_{1, 1} A_{1, 2} A_{1, 3}
A_{2, 1} A_{2, 2} A_{2, 3}
A_{3, 1} A_{3, 2} A_{3, 3}
N
b_1
\vdots
b_N

Output

If we will have a bingo, print Yes; otherwise, print No.

Sample Input 1

84 97 66
79 89 11
61 59 7
7
89
7
87
79
24
84
30

Sample Output 1

Yes

We will mark A_{1, 1}, A_{2, 1}, A_{2, 2}, A_{3, 3}, and complete the diagonal from the top-left to the bottom-right.

Sample Input 2

41 7 46
26 89 2
78 92 8
5
6
45
16
57
17

Sample Output 2

No

We will mark nothing.

Sample Input 3

60 88 34
92 41 43
65 73 48
10
60
43
88
11
48
73
65
41
92
34

Sample Output 3

Yes

We will mark all the squares.

Solution in Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*

fun main(args: Array<String>) {
    val scn = Scanner(System.`in`)
    var A = Array(4, {IntArray(4)})
    for(i in 1..3){
        for(j in 1..3) {
            A[i][j] = scn.nextInt()
        }
    }

    val N = scn.nextInt()
    var b = IntArray(N+1)
    for(i in 1..N){
        b[i] = scn.nextInt()
    }

    for(i in 1..3){
        for(j in 1..3) {
            for(k in 1..N){
                if(A[i][j] == b[k]) A[i][j] = 0
            }
        }
    }

    var result = false
    // 横
    for(i in 1..3){
        if(isBingo(A[i][1], A[i][2], A[i][3])) result = true
    }

    // 縦
    for(j in 1..3){
        if(isBingo(A[1][j], A[2][j], A[3][j])) result = true
    }

    // 斜め
    if(isBingo(A[1][1], A[2][2], A[3][3])) result = true
    if(isBingo(A[1][3], A[2][2], A[3][1])) result = true

    if(result) {
        println("Yes")
    }else{
        println("No")
    }
}

fun isBingo(i:Int,j:Int,k:Int): Boolean {
    if(i==0 && j ==0 && k ==0) return true
    return false
}