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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>

typedef long long int int64;

#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define ABS(a) ((a)>(0)?(a):-(a))

#define POS(i,j)((i)*3+(j))

void run(void){
  int c[9];
  int i;
  for(i=0;i<9;i++) scanf("%d",c+i);

  for(i=1;i<3;i++){
    int t=c[POS(i,0)]-c[POS(0,0)];
    int j;
    for(j=1;j<3;j++){
      if(c[POS(i,j)]-c[POS(0,j)]!=t) break;
    }
    if(j<3){
      printf("No\n");
      return;
    }
  }

  for(i=1;i<3;i++){
    int t=c[POS(0,i)]-c[POS(0,0)];
    int j;
    for(j=1;j<3;j++){
      if(c[POS(j,i)]-c[POS(j,0)]!=t) break;
    }
    if(j<3){
      printf("No\n");
      return;
    }
  }
  printf("Yes\n");
  return;
}

int main(void){
  run();
  return 0;
}