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()
#define AND(x,y,z) set_intersection((x).begin(),(x).end(),(y).begin(),(y).end(),inserter((z),(z).end()))
#define CONCAT(x,y) (x).insert((x).end(),(y).begin(),(y).end())
#define DUMP(xs) for (auto x:xs) cout<<x<<' ';cout<<endl
#define FOR(i,a,b) for (int i=(int)(a);i<(int)(b);++i)
#define OR(x,y,z) set_union((x).begin(),(x).end(),(y).begin(),(y).end(),inserter((z),(z).end()))
#define OUT(x) cout<<x<<endl
#define REP(i,n) FOR(i,0,n)

typedef long long ll;
typedef pair<ll, ll> P;
typedef vector<int> VECINT;
typedef vector<ll> VECLL;

int main() {
  VECINT CS(9);
  REP(i,9) cin>>CS[i];
  int max_a1 = min(CS[0],CS[1]);
  max_a1 = min(max_a1,CS[2]);
  REP(a1,max_a1+1) {
    int b1 = CS[0]-a1;
    int b2 = CS[1]-a1;
    int b3 = CS[2]-a1;
    int a2 = CS[3]-b1;
    int a3 = CS[6]-b1;
    if (CS[4]==a2+b2 && CS[5]==a2+b3 && CS[7]==a3+b2 && CS[8]==a3+b3) {
      OUT("Yes");
      return 0;
    }
  }
  OUT("No");

  return 0;
}