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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
public class Main {
	
	public static void main(String[] args) {
		Scanner stdIn = new Scanner(System.in);
		int[][] c = new int[3][3];
		for(int i = 0; i < 3; i++) {
			for(int j = 0; j < 3; j++) {
				c[i][j] = stdIn.nextInt();
			}
		}
		
		/*
		int s1 = c[0][0]-c[1][0];
		int s2 = c[0][1]-c[1][1];
		int s3 = c[0][2]-c[1][2];
		
		int s4 = c[1][0]-c[2][0];
		int s5 = c[1][1]-c[2][1];
		int s6 = c[1][2]-c[2][2];
		
		int s7 = c[0][0]-c[0][1];
		int s8 = c[1][0]-c[1][1];
		int s9 = c[2][0]-c[2][1];
		
		int s10 = c[0][1]-c[0][2];
		int s11 = c[1][1]-c[1][2];
		int s12 = c[2][1]-c[2][2];
		
		if(s1 == s2 && s2 == s3 && s4 == s5 && s5 == s6 && s7 == s8 && s8 == s9 && s10 == s11 && s11 == s12) {
			System.out.println("Yes");
		}
		else {
			System.out.println("No");
		}
		*/
		
		int[] a = new int[3];
		int[] b = new int[3];
		a[0] = 0;
		b[0] = c[0][0];
		b[1] = c[0][1];
		b[2] = c[0][2];
		a[1] = c[1][0]-b[0];
		a[2] = c[2][0]-b[0];
		
		boolean flag = true;
		for(int i = 0; i < 3; i++) {
			for(int j = 0; j < 3; j++) {
				if(c[i][j] != a[i] + b[j]) {
					flag = false;
					break;
				}
			}
		}
		
		if(flag) {
			System.out.println("Yes");
		}
		else {
			System.out.println("No");
		}
	}

}