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: Wrong Answer)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = 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] = Integer.parseInt(sc.next());
		int a1min = Math.min(Math.min(c[0][0], c[0][1]), Math.min(c[0][1], c[0][2]));
		int a2min = Math.min(Math.min(c[1][0], c[1][1]), Math.min(c[1][1], c[1][2]));
		int a3min = Math.min(Math.min(c[2][0], c[2][1]), Math.min(c[2][1], c[2][2]));
		int b1min = Math.min(Math.min(c[0][0], c[1][0]), Math.min(c[1][0], c[2][0]));
		int b2min = Math.min(Math.min(c[0][1], c[1][1]), Math.min(c[1][1], c[2][1]));
		int b3min = Math.min(Math.min(c[0][2], c[1][2]), Math.min(c[1][2], c[2][2]));
		for (int i = 0 ; i < 3; i++)
			c[0][i] -= a1min;
		for (int i = 0 ; i < 3; i++)
			c[1][i] -= a2min;
		for (int i = 0 ; i < 3; i++)
			c[2][i] -= a3min;
		for (int i = 0 ; i < 3; i++)
			c[i][0] -= b1min;
		for (int i = 0 ; i < 3; i++)
			c[i][1] -= b2min;
		for (int i = 0 ; i < 3; i++)
			c[i][2] -= b3min;
		boolean yes = true;
		for (int i = 0; i < 3; i++)
			for (int j = 0; j < 3; j++)
				if (c[i][j] != 0)
					yes = false;
		System.out.println((yes)? "Yes" : "No");
	}
}