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<cstdio>
#include<bits/stdc++.h>
typedef long long int ll;
#define REP(i, n) for(int i = 0; i < (n); i++)
#define FOR_IN(i, a, b) for(int i = (a); i < (b); i++)
#define BETWEEN(x, a, b) ((x) >= (a) && (x) <= (b))
#define BIT(b, i) (((b) >> (i)) & 1)
#define LOG_F 1
#define LOG(...) if(LOG_F) fprintf(stderr, __VA_ARGS__)

using namespace std;

ll pow(int x, int n){
  return n == 0 ? 1 : x * pow(x, n - 1);
}


/*

 */
int c[3][3];

int main(){
  REP(i,3){
    REP(j,3){
      cin>>c[i][j];
    }
  }

  REP(i,3){
    FOR_IN(j,1,3){
      c[i][j] -= c[i][0];
    }
  }
  
  FOR_IN(i,1,3){
    FOR_IN(j,0,3){
      c[i][j] -= c[0][j];
    }
  }

  int f=true;

  FOR_IN(i,1,3){
    FOR_IN(j,1,3){
      f = f && c[i][j] == 0;
    }
  }

  if(f)
    cout<<"Yes\n";
  else
    cout<<"No\n";

  return 0;
}