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){
    int c0[3],c1[3],c2[3],i,temp[3];
    scanf("%d %d %d",&c0[0],&c0[1],&c0[2]);
    scanf("%d %d %d",&c1[0],&c1[1],&c1[2]);
    scanf("%d %d %d",&c2[0],&c2[1],&c2[2]);
    
    for(i=0;i<3;i++){
        temp[i] = c1[i]-c0[i];
    }
    if(temp[0] != temp[1] || temp[1] != temp[2]){
        printf("No\n");
        return 0;
    }
    
    for(i=0;i<3;i++){
        temp[i] = c2[i]-c1[i];
    }
    if(temp[0] != temp[1] || temp[1] != temp[2]){
        printf("No\n");
        return 0;
    }
    temp[0] = c0[1]-c0[0];
    temp[1] = c1[1]-c1[0];
    temp[2] = c2[1]-c2[0];
    if(temp[0] != temp[1] || temp[1] != temp[2]){
        printf("No\n");
        return 0;
    }
    temp[0] = c0[2]-c0[1];
    temp[1] = c1[2]-c1[1];
    temp[2] = c2[2]-c2[1];
    if(temp[0] != temp[1] || temp[1] != temp[2]){
        printf("No\n");
        return 0;
    }
    printf("Yes\n");
    return 0;
}