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<iostream>
using namespace std;

int main(){
    int c[10][10];
    int d[10][10];
    int i, j, a = 0;
    for(i = 1; i < 4; i++){
        for(j = 1; j < 4; j++){
            cin >> c[i][j]; //入れる
        }
    }
    for(i = 1; i < 4; i++){
        c[i][4] = c[i][1]; //右端の操作
    }
    for(i = 1; i < 4; i++){
        for(j = 1; j < 4; j++){
            d[i][j] = c[i][j] - c[i][j+1]; //差をとる
        }
    }
    if(d[1][1] == d[2][1] && d[2][1] == d[3][1] && d[1][2] == d[2][2] && d[2][2] == d[3][2] && d[1][3] == d[2][3] && d[2][3] == d[3][3]){
        a = 1;
    }

    for(i = 1; i < 4; i++){
        c[4][i] = c[1][i];
    }
    for(i = 1; i < 4; i++){
        for(j = 1; j < 4; j++){
            d[i][j] = c[i][j] - c[i+1][j];
        }
    }
    if(d[1][1] == d[1][2] && d[1][2] == d[1][3] && d[2][1] == d[2][2] && d[2][2] == d[2][3] && d[3][1] == d[3][2] && d[3][2] == d[3][3]){
        a = 1;
    }
    if(a == 1){
        cout << "Yes" << endl;
    }
    else{
        cout << "No" << endl;
    }
    return 0;
}