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 c[3][3];

int main() {

	int i,j;

	for(i=0;i<3;i++){
		for(j=0;j<3;j++){
			scanf(" %d",&c[i][j]);
		}
	}

	for(i=0;i<=c[0][0];i++){
		int b1=c[0][0]-i;
		int b2=c[0][1]-i;
		int b3=c[0][2]-i;
		if(0<=b1&&b1<=100 && 0<=b2&&b2<=100 && 0<=b3&&b3<=100){
			int a2_x=c[1][0]-b1;
			int a2_y=c[1][1]-b2;
			int a2_z=c[1][2]-b3;
			if(a2_x==a2_y&&a2_x==a2_z &&0<=a2_x&&a2_x<=100){
				int a3_x=c[2][0]-b1;
				int a3_y=c[2][1]-b2;
				int a3_z=c[2][2]-b3;
				if(a3_x==a3_y&&a3_x==a3_z &&0<=a3_x&&a3_x<=100){
					printf("Yes\n");
					return 0;
				}
			}
		}
	}

	printf("No\n");

	return 0;
}