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 C (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>

int main(void)
{
  int bingo[9];
  int n;
  int b[100];
  int i, j, k;
  int result;

  // 入力
  scanf("%d %d %d", &bingo[0], &bingo[1], &bingo[2]);
  scanf("%d %d %d", &bingo[3], &bingo[4], &bingo[5]);
  scanf("%d %d %d", &bingo[6], &bingo[7], &bingo[8]);
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &b[i]);
  }

  // bingo列
  int bingoArray[][3] = {
    {0, 1, 2},
    {3, 4, 5},
    {6, 7, 8},
    {0, 3, 6},
    {1, 4, 7},
    {2, 5, 8},
    {0, 4, 8},
    {2, 4, 6}
  };

  result = 0;
  // bingoArrayの外側
  for (i = 0; i < 8; i++) {
    // bingoArrayの内側
    for (j = 0; j < 3; j++) {
      // bを回す
      for (k = 0; k < n; k++) {
        if (bingo[bingoArray[i][j]] == b[k]) {
          result += 1;
          break;
        }
      }
    }
    if (result == 3) {
      break;
    } else {
      result = 0;
    }
  }

  if (result == 3) {
    printf("Yes\n");
  } else {
    printf("No\n");
  }

  return 0;
}