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>
#include <stdbool.h>
#define LEN 3

int are_all_positive(int a[LEN], int b[LEN])
{
	return ((a[0] >= 0) && (a[1] >= 0) && (a[2] >= 0) && (b[0] >= 0) && (b[1] >= 0) && (b[2] >= 0));
}

int is_matched(int a[LEN], int b[LEN], int c[LEN][LEN])
{
	return ((c[0][1] == a[0] + b[1]) && (c[0][2] == a[0] + b[2]) && (c[1][2] == a[1] + b[2]) && (c[2][0] == a[2] + b[0]));
}

int main()
{
	int c[LEN][LEN];
	int a[LEN], b[LEN];
	bool is_correct = false;

	for (int i = 0; i < LEN; i++) {
		scanf("%d %d %d", &c[i][0], &c[i][1], &c[i][2]);
	}

	a[0] = 0;
	while (1) {
		if (a[0] > c[0][0]) goto RESULT;
		b[0] = c[0][0] - a[0];
		a[1] = c[1][0] - b[0];
		b[1] = c[1][1] - a[1];
		a[2] = c[2][1] - b[1];
		b[2] = c[2][2] - a[2];

		if (!is_matched(a, b, c)) goto INCREMENT;
		if (are_all_positive(a, b)) {
			is_correct = true;
			goto RESULT;
		}
		INCREMENT:a[0]++;
	}

	RESULT:if (is_correct == true) {
		printf("Yes\n");
	} else {
		printf("No\n");
	}
	return 0;
}