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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args : Array<String>) {
    val cc = Array(3) { readLine()!!.split(" ").map { it.toInt() } }
    val ansA = mutableListOf(0, 0, 0)
    val ansB = mutableListOf(0, 0, 0)

    ansB[0] = cc[0][0]
    ansB[1] = cc[0][1]
    ansB[2] = cc[0][2]

    ansA[1] = cc[1][0] - ansB[0]
    ansA[2] = cc[2][0] - ansB[0]

    cc.forEachIndexed { idx, row ->
        repeat(3) {
            if (row[it] != ansA[idx] + ansB[it]) {
                println("No")
                return
            }
        }
    }

    println("Yes")
}