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 ll long long
#define MODV 1000000007
#define INF INT_MAX // 2147483647
#define INFLL LLONG_MAX // 9223372036854775807
#define rep(i, n) for(ll i=0, i##_len=(ll)(n); i<i##_len; i++)
#define repf(i, n) for(ll i=1, i##_len=(ll)(n+1); i<i##_len; i++)
#define per(i, n) for(ll i=((ll)(n))-1; i>=0; i--)
#define perf(i, n) for(ll i=((ll)(n)); i>0; i--)
#define all(v) v.begin(), v.end()
#define endl "\n"
#define vi vector<ll>
#define vvi vector<vector<ll>>
#define Yes() cout << "Yes" << endl
#define YES() cout << "YES" << endl
#define No() cout << "No" << endl
#define NO() cout << "NO" << endl
#define Init() std::ios::sync_with_stdio(false); std::cin.tie(0); std::cout<<fixed<<setprecision(15);
template<class T>bool chmax(T &a, const T &b){ if(a<b){ a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b){ if(b<a){ a=b; return 1; } return 0; }
using namespace std;

int main(){
    Init();

    // 入力処理
    vector<vector<ll>> c(3, vector<ll>(3, 0));
    rep(i, 3) rep(j, 3) cin >> c[i][j];

    vi a(3), b(3);
    for(a[0]=-100; a[0]<=100; a[0]++){
        b[0] = c[0][0]-a[0];
        b[1] = c[0][1]-a[0];
        b[2] = c[0][2]-a[0];
        a[1] = c[1][0]-b[0];
        a[2] = c[2][0]-b[0];
        bool flag = false;
        rep(i, 3){
            rep(j, 3){
                if(c[i][j] != a[i]+b[j]){
                    flag = true;
                    break;
                }
            }
            if(flag) break;
        }
        if(flag) continue;
        Yes();
        return 0;
    }   
    No();
}