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 Go (1.6) (judge verdict: Wrong Answer)

package main

import "fmt"

func main(){
	var A [3][3] int
	var N int
	var a [3][3] bool

	for i := 0 ; i < 3 ; i++ {
		for j := 0 ; j < 3 ; j++ {
			fmt.Scan(&A[i][j])
		}
	}
	fmt.Scan(&N)

	b := make([]int, N)
	
	for i := 0 ; i < N ; i++ {
		fmt.Scan(&b[i])
	}

	for k := 0 ; k < N ; k++ {
		for i := 0 ; i < 3 ; i++ {
			for j := 0 ; j < 3 ; j++{
				if A[i][j] == b[k] {
					a[i][j] = true
				}
			}
		}
	}
	var tate, yoko [3] int
	var naname  int
	for i := 0 ; i < 3 ; i++ {
		if a[i][0] == true {
			tate[i]++
		}
		if a[0][i] == true {
			yoko[i]++
		}
	}
	if (a[0][0] == true && a[1][1] == true && a[2][2] == true) || (a[0][2] == true && a[1][1] == true && a[2][0] == true) {
		naname++
	}

	if tate[0] == 3 || tate[1] == 3 || tate[2] == 3 || yoko[0] == 3 || yoko[1] == 3 || yoko[2] == 3 || naname > 0 {
		fmt.Println("Yes")
	}else {
		fmt.Println("No")
	}
	//fmt.Println(a)
	// for i := 0 ; i < 3 ; i++ {
	// 	for j := 0 ; j < 3 ; j++ {
	// 		fmt.Print(a[i][j])
	// 	}
	// 	fmt.Println()
	// }
}