Problem:
Score: 300 points

Problem Statement

We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left.

According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j.

Determine if he is correct.

Constraints

c_{i, j} \ (1 \leq i \leq 3, 1 \leq j \leq 3) is an integer between 0 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

c_{1,1} c_{1,2} c_{1,3}
c_{2,1} c_{2,2} c_{2,3}
c_{3,1} c_{3,2} c_{3,3}

Output

If Takahashi's statement is correct, print Yes; otherwise, print No.

Sample Input 1

1 0 1
2 1 2
1 0 1

Sample Output 1

Yes

Takahashi is correct, since there are possible sets of integers such as: a_1=0,a_2=1,a_3=0,b_1=1,b_2=0,b_3=1.

Sample Input 2

2 2 2
2 1 2
2 2 2

Sample Output 2

No

Takahashi is incorrect in this case.

Sample Input 3

0 8 8
0 8 8
0 8 8

Sample Output 3

Yes

Sample Input 4

1 8 6
2 9 7
0 7 7

Sample Output 4

No

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int main(void)
{
	//変数を定義。
	int c[3][3];
	int flag = 0;
 
	//3×3のグリッドを入力。
	for (int i = 0; i < 3; i++) {
		scanf("%d %d %d\n", &c[i][0], &c[i][1], &c[i][2]);
	}
 
	/*表の上の数値から下の数値を引いた値がそれぞれ等しいと条件が満たされる。
	  例えば、c[0][0] - c[1][0] と c[0][1] - c[1][1]の差が等しい。*/
	for (int j = 0; j < 2; j++) {
		for (int k = 0; k < 2; k++) {
			if (c[j][k] - c[j + 1][k] != c[j][k + 1] - c[j + 1][k + 1]) {
				printf("No\n");
                return 0;
			}
		}
	}
		printf("Yes\n");
		
		return 0;
}