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: Accepted)

#include "bits/stdc++.h"

#define debug(x) cout<<#x<<": "<<x<<endl
#define rep(i,n) for (int i=0;i<(n);i++)
#define FOR(i,a,b) for (int i=(a);i<=(b);i++)
#define all(a) (a).begin(),(a).end()
using namespace std;
typedef vector<int> VI;
typedef vector<vector<int>> VVI;
typedef long long ll;



void solve() {
	VI v(9);
	rep(i, 9) cin >> v[i];
	
	bool flag = true;
	rep(j, 2) {
		if (v[3 + j * 3] - v[j * 3] != v[4 + j * 3] - v[1 + j * 3] || v[4 + j * 3] - v[1 + j * 3] != v[5 + j * 3] - v[2 + j * 3]) {
			flag = false;
		}
	}
	rep(j, 2) {
		if (v[j + 1] - v[j] != v[j + 4] - v[j + 3] || v[j + 4] - v[j + 3] != v[j + 7] - v[j + 6]) {
			flag = false;
		}
	}
	cout << (flag ? "Yes" : "No") << endl;
}


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	solve(); 
	return 0;
}