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 rep(i,n) for(long long int (i)=0;(i)<(int)(n);(i)++)
#define rrep(i,a,b) for(long long int i=(a);i<(b);i++)
#define rrrep(i,a,b) for(long long int i=(a);i>=(b);i--)
#define all(v) (v).begin(), (v).end()
#define pb(q) push_back(q)
#define Abs(a,b) max(a,b)-min(a,b)
#define YES(condition) if(condition){cout << "YES" << endl;}else{cout << "NO" << endl;}
#define Yes(condition) if(condition){cout << "Yes" << endl;}else{cout << "No" << endl;}
#define Cout(x) cout<<(x)<<endl
#define POSSIBLE(condition) if(condition){cout << "POSSIBLE" << endl;}else{cout << "IMPOSSIBLE" << endl;}
#define Possible(condition) if(condition){cout << "Possible" << endl;}else{cout << "Impossible" << endl;}
#define possible(condition) if(condition){cout << "possible" << endl;}else{cout << "impossible" << endl;}
#define Size(n) (n).size()

typedef long long ll;


using namespace std;

const int INF = 1e9,MOD = 1e9 + 7,ohara=1e6;
const ll LINF = 1e18;

/*---------------------------------------------------------------
long long int kaizyo(long long int hh){
    cmp=1;
    while(hh>1){
        cmp=(cmp*hh)%MOD;
        hh--;
    }
    return cmp;
}
long long int ruizyo(long long int aa, long long int bb){
    if(aa==0){
        return 1;
    }
    else if(aa%2==0){
        long long int tt=ruizyo(aa/2,bb);
        return (tt*tt)%MOD;
    }
    else{
        return (ruizyo(aa-1,bb)*bb)%MOD;
    }
}フェルマ－のア
---------------------------------------------------------------

while(x!=0){
            sum+=x%10;
          / x/=10;
        }
        各桁の和
---------------------------------------------------------------

pair<int,int> p[100000];
cin >> tmp;
p[i]=make_pair(tmp,i);
cout << p[i].second+1 << endl;//ペアの右側つまりiを出力

---------------------------------------------------------------

bool f[100001];//1000000以下の素数を調べよう！
rrep(i,2,100001){
       f[i]=false;
    }
rrep(i,2,100001){
        if(!f[i]){
            for(int j=i+i;j<=100000;j+=i){
                f[j]=true;
            }
        }
    }
    for(int i=3;i<=100000;i+=2){
        if(!f[i]){
            c[i]++;
        }
    }

---------------------------------------------------------------

long long gcd(long long aaa,long long bbb){
    if(bbb==0){
        return aaa;
    }
    return gcd(bbb,aaa%bbb);
}
 
long long lcm(long long aaa,long long bbb){
    long long g = gcd(aaa,bbb);
    return aaa/g * bbb;
}左から最大公約数と最小公倍数


---------------------------------------------------------------

long long int prime_cnt[10000];
       for(int i=2;i*i<=n;i++){
           while(n%i==0){
               n/=i;
               prime_cnt[i]+=1;
           }
           if(n>1){
               prime_cnt[n]+=1;
               break;
           }
      }ある数nを素因数分解しましょう

---------------------------------------------------------------
debug:
cout<<"i: "<<i<<" j: "<<j<<" cnt: "<<cnt<<"\n";

---------------------------------------------------------------*/

long long int n,cnt=0,ans=0,a[ohara],b,c,d,cmp[5][5],cmpp,m,h,w,x,y,sum=0,pos;
int dy[]={1,0,-1,0};
int dx[]={0,1,0,-1};
string alph("abcdefghijklmnopqrstuvwxyz"),s;
bool fl=true;
struct edge{int to,cost;};

//-------------------------↓↓↓↓↓↓------------------------

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

        rep(i,3){
            rep(j,3){
                cin>>cmp[i][j];
            }
        }
        if(cmp[0][0]-cmp[1][0]==cmp[0][1]-cmp[1][1]&&cmp[0][1]-cmp[1][1]==cmp[0][2]-cmp[1][2]){
            if(cmp[1][0]-cmp[2][0]==cmp[1][1]-cmp[2][1]&&cmp[1][1]-cmp[2][1]==cmp[1][2]-cmp[2][2]){
                Cout("Yes");
                return 0;
            }
        }
        Cout("No");
       
return 0;
}