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 <bits/stdc++.h>
using namespace std;
int main()
{
    int n = 3;
    vector<vector<int>> c(n, vector<int>(n));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> c.at(i).at(j);
        }
    }

    for (int i = 0; i < n -1; i++)
    {
        for (int j = 0; j < n - 1; j++)
        {
            if (c.at(i+1).at(j) - c.at(i).at(j) != c.at(i+1).at(j+1) - c.at(i).at(j+1))
            {
                cout << "No" << endl;
                return 0;
            }
        }
    }

    cout << "Yes" << endl;

    return 0;
}