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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

#define rep(i, n) REP(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define MSG(a) cout << #a << " " << a << endl;
#define REP(i, x, n) for (int i = x; i < n; i++)
#define OP(m) cout << m << endl
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

int main()
{
    int c[3][3];
    rep(i, 3) rep(j, 3) cin >> c[i][j];
    rep(i, 101)
    {
        rep(j, 101)
        {
            rep(k, 101)
            {
                int b1 = c[0][0] - i;
                int b2 = c[0][1] - i;
                int b3 = c[0][2] - i;
                if (b1 == c[1][0] - j && b1 == c[2][0] - k && b2 == c[1][1] - j && b2 == c[2][1] - k && b3 == c[1][2] - j && b3 == c[2][2] - k)
                    if (0 <= b1 && b1 <= 100 && 0 <= b2 && b2 <= 100 && 0 <= b3 && b3 <= 100)
                    {
                        OP("Yes");
                        return 0;
                    }
            }
        }
    }

    OP("No");
    return 0;
}