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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int c[3][3];

int main(int argc, char* argv) {

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

        const int c00 = c[0][0];
        const int c01 = c[0][1];
        const int c02 = c[0][2];
        // start search
        for (int a1 = 0; a1 <= c00; a1++) {
                for (int a2 = 0; a2 <= c01; a2++) {
                        for (int a3 = 0; a3 <= c02; a3++) {
                                const int b1 = c00 - a1;
                                const int b2 = c01 - a2;
                                const int b3 = c02 - a3;

                                const int a2t = c[1][0] - b1;
                                const int a3t = c[2][0] - b1;
                                if (
                                  (a2t == (c[1][1] - b2)) && (a2t == (c[1][2] - b3))
                                && (a3t == (c[2][1] - b2)) && (a3t == (c[2][2] - b3))) {
                                        printf("Yes\n");
                                        exit(0);
                                }
                        }
                }
        }

        printf("No\n");
        return 0;
}