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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
#include <stack>

using ll = long long;

int main() {

    std::vector<std::vector<int>> c(3, std::vector<int>(3));
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 3; j++) {
            std::cin >> c[i][j];
        }
    }

    std::vector<int> a(3), b(3);
    for (int i = 0; i <= c[0][0]; i++) {
        a[0] = i;
        int b1 = c[0][0] - i;
        if (b1 < 0)continue;
        b[0] = b1;
        int b2 = c[0][1] - i;
        if (b2 < 0)continue;
        b[1] = b2;
        int b3 = c[0][2] - i;
        if (b3 < 0)continue;
        b[2] = b3;
        int a2 = c[1][0] - b1;
        if (a2 < 0)continue;
        if (a2 + b2 != c[1][1])continue;
        if (a2 + b3 != c[1][2])continue;
        int a3 = c[2][0] - b1;
        if (a3 < 0)continue;
        if (a3 + b2 != c[2][1])continue;
        if (a3 + b3 != c[2][2])continue;
        std::cout << "Yes" << std::endl;
        return 0;
    }

    std::cout << "No" << std::endl;
    return 0;
}