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: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;
int main(void){
    // Your code here!
    long S[3][3];
    cin>>S[0][0]>>S[0][1]>>S[0][2];
    cin>>S[1][0]>>S[1][1]>>S[1][2];
    cin>>S[2][0]>>S[2][1]>>S[2][2];
    
    long i,j,k,l,m,n;
    long ii,jj,kk,ll,mm,nn;
    ii=min(min(S[0][0]+1,S[0][1]+1),S[0][2]+1);
    jj=min(min(S[1][0]+1,S[1][1]+1),S[1][2]+1);
    kk=min(min(S[2][0]+1,S[2][1]+1),S[2][2]+1);
    ll=min(min(S[0][0]+1,S[1][0]+1),S[2][0]+1);
    mm=min(min(S[0][1]+1,S[1][1]+1),S[2][1]+1);
    nn=min(min(S[0][2]+1,S[1][2]+1),S[2][2]+1);
    
    
    
    for (i=0;i<=ii;i++)
    {
        for (j=0;j<=jj;j++)
        {
            for (k=0;k<=kk;k++)
            {
                for (l=0;l<=ll;l++)
                {
                    for (m=0;m<=mm;m++)
                    {
                        for (n=0;n<=nn;n++)
                        {
                            
                            if(S[0][0]==i+l && S[1][0]==j+l && S[2][0]==k+l && S[0][1]==i+m && S[1][1]==j+m && S[2][1]==k+m && S[0][2]==i+n && S[1][2]==j+n && S[2][2]==k+n)
                            {
                                cout<<"yes"<<endl;
                                return 0;
                            
                                
                            }
                        }
                    }
                }
            }
        }
    }
    
    
    
    
    
    
    
    
    
    
    
    cout<<"No"<<endl;
    
}