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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using static System.Console;
using static System.Math;

public class Hello{
    public static void Main(){
//        int kazu = int.Parse(ReadLine());
        string[] input0 = ReadLine().Split(' ');
        string[] input1 = ReadLine().Split(' ');
        string[] input2 = ReadLine().Split(' ');
        int[,] suuji = new int[3,3];
        for(int i=0;i<3;i++){
            suuji[0,i] = int.Parse(input0[i]);
            suuji[1,i] = int.Parse(input1[i]);
            suuji[2,i] = int.Parse(input2[i]);
        }
/*        for(int i=0;i<3;i++){
            for(int j=0;j<3;j++){
                Write(suuji[i,j]);
            }
            WriteLine("");
        }*/
        int count = 0;
        for(int a1=0;a1<=suuji[0,0];a1++){
            for(int a2=0;a2<=suuji[1,0];a2++){
                for(int a3=0;a3<=suuji[2,0];a3++){
                    int[] b = new int[3];
                    for(int i=0;i<3;i++){
                        b[i] = suuji[0,i] -a1;
                        if(b[i] != suuji[1,i]-a2 || b[i] != suuji[2,i]-a3){
                            count = 0;
                            break;
                        }else{
                            count++;
                        }
                    }
                    if(count > 0){ break; }
                }
                if(count > 0){ break; }
            }
            if(count > 0){ break; }
        }
        if(count > 0){
            WriteLine("Yes");
        }else{
            WriteLine("No");
        }
    }
}