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)

#define _CRT_SECURE_NO_WARNINGS

#include<iostream>
#include<string>
#include<vector>
#include<list>
#include<stack>
#include<queue>
#include<map>
#include<tuple>
#include<algorithm>
#include<numeric>
#include<math.h>

#define REPS(i, a, b) for (int i = (a); i < (b); i++)
#define REP(i, n) for (int i = 0; i < (n); i++)

using namespace std;

using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;

const int mod = 1e9 + 7;
const long double pi = 3.141592653589793238462643383279502884197;
const long double eps = 1e-7;

int main() {
	int c[3][3];
	REP(i, 3) {
		REP(j, 3) {
			cin >> c[i][j];
		}
	}

	bool col = false, row = false;

	int a12 = c[1][0] - c[0][0];
	int a13 = c[2][0] - c[0][0];
	int b12 = c[0][1] - c[0][0];
	int b13 = c[0][2] - c[0][0];

	if(a12 == (c[1][1]-c[0][1]) && a12 == (c[1][2]-c[0][2]) 
		&& a13 == (c[2][1] - c[0][1]) && a13 == (c[2][2] - c[0][2])) {
		col = true;
	}

	if (b12 == (c[1][1] - c[1][0]) && b12 == (c[2][1] - c[2][0])
		&& b13 == (c[1][2] - c[1][0]) && b13 == (c[2][2] - c[2][0])) {
		row = true;
	}

	if (col && row) {
		cout << "Yes";
	}
	else {
		cout << "No";
	}
	

	system("pause");
	return 0;
}