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>
using namespace std;
#define all(x) (x).begin(),(x).end()
using ll = long long;
#define vl vector<long long>
#define vvl vector<vector<long long>>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
using pll = pair<long long,long long>;
#define cY cout<<"Yes"<<endl
#define cN cout<<"No"<<endl
const long long INF = 1LL << 60;
const long long MOD = 1000000007;

//最大最小の入れ替え
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

//約数列挙
vector<long long> divisor(long long n) {
    vector<long long> ret;
    for (long long i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            ret.push_back(i);
            if (i * i != n) ret.push_back(n / i);
        }
    }
    sort(ret.begin(), ret.end()); // 昇順に並べる
    return ret;
}
//以上テンプレ
int main(){
    vvl vec(3,vl(3));
    rep(i,3){
        rep(j,3){
            cin>>vec.at(i).at(j);
        }
    }
    rep(i,2){
        if(vec.at(i).at(0)-vec.at(i).at(1)!=vec.at(i+1).at(0)-vec.at(i+1).at(1)){
            cN;
            return 0;
        }else if(vec.at(i).at(1)-vec.at(i).at(2)!=vec.at(i+1).at(1)-vec.at(i+1).at(2)){
            cN;
            return 0;
        }
    }
    cY;
    }