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 Go (1.6) (judge verdict: Wrong Answer)

package main

import "fmt"

func main() {
	c := [][]int{
		{0, 0, 0},
		{0, 0, 0},
		{0, 0, 0},
	}

	for i := 0; i < 3; i++ {
		for j := 0; j < 3; j++ {
			fmt.Scan(&c[i][j])
		}
	}

	b := make([]int, 3)
	for n := 0; n <= 100; n++ {
		for i := 0; i < 3; i++ {
			b[i] = c[0][i] - n
		}
		if b[0] < 0 || b[1] < 0 || b[2] < 0 {
			break
		}

		for i := 0; i < 3; i++ {
			if c[1][i] < b[i] {
				fmt.Println("No")
				return
			}
		}

		for i := 0; i < 3; i++ {
			if c[2][i] < b[i] {
				fmt.Println("No")
				return
			}
		}

		break
	}

	for n := 0; n <= 100; n++ {
		for i := 0; i < 3; i++ {
			b[i] = c[1][i] - n
		}
		if b[0] < 0 || b[1] < 0 || b[2] < 0 {
			break
		}

		for i := 0; i < 3; i++ {
			if c[0][i] < b[i] {
				fmt.Println("No")
				return
			}
		}

		for i := 0; i < 3; i++ {
			if c[2][i] < b[i] {
				fmt.Println("No")
				return
			}
		}

		break
	}

	for n := 0; n <= 100; n++ {
		for i := 0; i < 3; i++ {
			b[i] = c[2][i] - n
		}
		if b[0] < 0 || b[1] < 0 || b[2] < 0 {
			break
		}

		for i := 0; i < 3; i++ {
			if c[0][i] < b[i] {
				fmt.Println("No")
				return
			}
		}

		for i := 0; i < 3; i++ {
			if c[1][i] < b[i] {
				fmt.Println("No")
				return
			}
		}

		break
	}
	fmt.Println("Yes")

}