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 <map>
#include <vector>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <climits>
#include <stack>
#include <queue>
#include <set>
#include <cctype>
#include <bitset>
#include <type_traits>

#include <list>
using namespace std;

#define REP(i, n) for (int i = 0; i < (n); ++i)
#define rep(i, a, b) for (int i = a; i < (b); ++i)
#define YES(j) cout << (j ? "YES" : "NO") << endl;
#define Yes(j) std::cout << (j ? "Yes" : "No") << endl;
#define yes(j) std::cout << (j ? "yes" : "no") << endl;

int main(void)
{
    int c[3][3];
    REP(i, 3)
    {
        REP(j, 3)
        {
            cin >> c[i][j];
        }
    }
    int a[3][3], b[3][3];
    // a1 = i
    // b1 = j
    for (int i = 0; i <= 100; i++)
    {
        bool flag = true;
        int j = c[0][0] - i;
        a[0][0] = i;
        b[0][0] = j;
        // c21 = a2 + b1
        // c31 = a3 + b1
        for (int k = 1; k <= 2; k++)
        {
            a[k][0] = -b[0][0] + c[k][0];
        }
        // c12 = a1 + b2
        // c13 = a1 + b3
        for (int k = 1; k <= 2; k++)
        {
            b[0][k] = -a[0][0] + c[0][k];
        }

        for (int k = 1; k <= 2; k++)
        {
            for (int l = 1; l <= 2; l++)
            {
                // ckl = ak + bl
                if (not(c[k][l] == a[k][0] + b[0][l]))
                {
                    flag = false;
                }
            }
        }
        if (flag)
        {
            Yes(1);
            return 0;
        }
    }
    Yes(0);
}