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>

#define REP(i, n) for(int (i)=0;(i)<(n);++(i))
#define MOD 1000000007
#define GRID 3

using namespace std;

typedef long long ll;

int main()
{
    int c[GRID][GRID] = {0};
    REP(i, GRID)
    {
        REP(j, GRID)
        {
            cin >> c[i][j];
        }
    }
    // a1, a2, a3 を固定してb1, b2, b3 が存在するか確かめる
    int a1, a2, a3;
    for (a1 = 0; a1 <= 100; ++a1)
    {
        for (a2 = 0; a2 <= 100; ++a2)
        {
            for (a3 = 0; a3 <= 100; ++a3)
            {
                if ((c[0][0] - a1 >= 0) && (c[0][0] - a1 == c[1][0] - a2) && (c[0][0] - a1 == c[2][0] - a3) &&
                    (c[0][1] - a1 >= 0) && (c[0][1] - a1 == c[1][1] - a2) && (c[0][1] - a1 == c[2][1] - a3) &&
                    (c[0][2] - a1 >= 0) && (c[0][2] - a1 == c[1][2] - a2) && (c[0][2] - a1 == c[2][2] - a3) )
                    {
                        cout << "Yes" << endl;
                        return 0;
                    }
            }
        }
    }
    cout << "No" << endl;
    return 0;
}