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 <math.h>
#include <string>
#include <algorithm>
#include <ctype.h>
#include <vector>
using namespace std;

#define ll long long int

int main()
{
    int c[3][3];
    int a[3], b[3];

    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < 3; j++)
        {
            cin >> c[i][j];
        }
    }
    for (int i = 0; i <= 100; i++)
    {
        for (int j = 0; j <= 100; j++)
        {
            for (int k = 0; k <= 100; k++)
            {
                a[0] = i;
                a[1] = j;
                a[2] = k;
                bool f = true;
                for (int l = 0; l < 3; l++)
                {
                    for (int m = 0; m < 3; m++)
                    {
                        if (l == 0)
                        {
                            b[m] = c[l][m] - a[l];
                        }
                        else
                        {
                            if (a[l] + b[m] != c[l][m])
                            {
                                f = false;
                                break;
                            }
                        }
                        /*if (a[0] == 0 && a[1] == 1 && a[2] == 0)
                        {
                            for (int l = 0; l < 3; l++)
                            {
                                cout << a[l] << " ";
                            }
                            for (int l = 0; l < 3; l++)
                            {
                                cout << b[l] << " ";
                            }
                            cout << endl;
                        }*/
                    }
                }
                if (f)
                {
                    cout << "Yes" << endl;
                    return 0;
                }
            }
        }
    }
    cout << "No" << endl;
}