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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

int main(int argc, const char *argv[]) {
  array<array<int, 3>, 3> c;
  for (int i = 0; i < 3; ++i) {
    for (int j = 0; j < 3; ++j) {
      cin >> c[i][j];
    }
  }

  bool ans = false;
  for (int a1 = 0; a1 <= 100 && !ans; ++a1) {
    array<int, 3> va, vb;
    va[0] = a1;
    vb[0] = c[0][0] - va[0];
    vb[1] = c[0][1] - va[0];
    vb[2] = c[0][2] - va[0];
    va[1] = c[1][0] - vb[0];
    va[2] = c[2][0] - vb[0];

    bool ok = true;
    for (int i = 0; i < 3 && ok; ++i) {
      for (int j = 0; j < 3 && ok; ++j) {
        if (c[i][j] == va[i] + vb[j]) {
          continue;
        }

        ok = false;
      }
    }

    if (ok) {
      ans = ok;
    }
  }

  cout << (ans ? "Yes" : "No") << '\n';
  return 0;
}