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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <string>
#include <queue>
#include <cmath>
#include <map>

using namespace std;
using ll = long long;

int main() {
	vector<vector<ll>> c(3, vector<ll>(3));
	for (ll i = 0; i < 3; i++)
	{
		for (ll j = 0; j < 3; j++)
		{
			cin >> c[i][j];
		}
	}
	for (ll i = 0; i < 2; i++)
	{
		if (c[1][i + 1] - c[1][i] == c[0][i + 1] - c[0][i] == c[2][i + 1] - c[2][i] &&
			c[i + 1][0] - c[i][0] == c[i + 1][1] - c[i][1] == c[i + 1][2] - c[i][2]) continue;
	}
	cout << "Yes" << endl;
	return 0;
}