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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define lli long long int
#define uli unsigned long long int
#define rep(i, m, n) for (lli i = m; i < (n); i++)
#define repe(i, m, n) for (lli i = m; i <= (n); i++)
#define ALL(x) (x).begin(), (x).end()
#define SIZE(x) ((lli)(x).size())
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x))
#define _GLIBCXX_DEBUG
const lli INF = 2147483647;
const lli MINF = -2147483648;
const lli LINF = 9223372036854775807;
const lli MOD = 1000000007; //10^9+7
const double PI = acos(-1);
#define SORT(n) sort(n.begin(), n.end())
#define SORTR(n) sort(n.begin(), n.end(), greater<int>())
#define REV(n) reverse(n.begin(), n.end())
#define pb push_back
#define eb emplace_back
#define mp make_pair
#pragma GCC optimize("Ofast")
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vlli = vector<lli>;
using vvli = vector<vlli>;
using vs = vector<string>;
using vvs = vector<vs>;
using vb = vector<bool>;
using vvb = vector<vb>;
using ll = long long;
// vector入力
template <typename T>
istream &operator>>(istream &is, vector<T> &vec)
{
  for (T &x : vec)
    is >> x;
  return is;
}
// vector出力
template <typename T>
ostream &operator<<(ostream &os, vector<T> &vec)
{
  // os << '{';
  for (int i = 0; i < vec.size(); i++)
  {
    os << vec[i] << (i + 1 == vec.size() ? "" : "");
  }
  // os << '}';
  return os;
}
// aよりもbが大きいならばaをbで更新する(更新されたならばtrueを返す)
template <typename T>
bool chmax(T &a, const T &b)
{
  if (a < b)
  {
    a = b; // aをbで更新
    return true;
  }
  return false;
}
// aよりもbが小さいならばaをbで更新する(更新されたならばtrueを返す)
template <typename T>
bool chmin(T &a, const T &b)
{
  if (a > b)
  {
    a = b; // aをbで更新
    return true;
  }
  return false;
}

int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  lli a, b, c, h, n, w, ans = 0, count = 0, count1 = 0;
  vector<pair<string, pair<lli, lli>>> pr;
  vlli d;
  rep(i, 0, 9)
  {
    cin >> a;
    d.pb(a);
    ans += a;
  }

  if (ans == 3 * (d[0] + d[4] + d[8]))
  {
    cout << "Yes" << endl;
  }
  else
  {
    cout << "No" << endl;
  }
}