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 <algorithm>
#include <cstdio>
#include <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
#include <bitset>
#include <cstring>
#include <deque>
using namespace std;
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)
#define CLR(mat) memset(mat, 0, sizeof(mat))
typedef long long ll;
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  int c[3][3];
  FOR(i,0,3) FOR(j,0,3) cin >> c[i][j];
  FOR(a1,-101,101) {
    FOR(a2,-101,101) {
      FOR(a3,-101,101) {
        int b1 = c[0][0] - a1;
        int b2 = c[0][1] - a1;
        int b3 = c[0][2] - a1;
        bool ok = true;
        if(c[1][0] != a2 + b1) ok = false;
        if(c[1][1] != a2 + b2) ok = false;
        if(c[1][2] != a2 + b3) ok = false;
        if(c[2][0] != a3 + b1) ok = false;
        if(c[2][1] != a3 + b2) ok = false;
        if(c[2][2] != a3 + b3) ok = false;
        if(ok) {
          cout << "Yes" << endl;
          return 0;
        }
      }
    }
  }
  cout << "No" << endl;
  return 0;
}