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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
#define INF 1e12

#define PB push_back
#define PF push_front
#define fi first
#define se second
#define pii pair<int, int>
#define pll pair<ll, ll>
#define vi vector<int>
#define vpi vector<pii>
#define vll vector<ll>
#define vpl vector<pll>
#define vvi vector<vector<int>>
#define vvl vector<vector<ll>>
#define MX(x) *max_element(all(x))
#define MN(x) *min_element(all(x))
#define ios ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define pr_d(x) cout << fixed << setprecision(15) << x << endl
#define ud(c, x) distance(c.begin(), upper_bound(all(c), x))
#define ld(c, x) distance(c.begin(), lower_bound(all(c), x))
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep2(i, a, b) for (int i = (a); i < (b); ++i)
#define rep3(i, n) for (int i = (n - 1); i >= 0; --i)
#define rep4(i, a, b) for (int i = (a); i > (b); --i)
#define pb push_back
#define out(x) cout << x << "\n"
bool odd(int i) { return i % 2; }

#define all(v) v.begin(), v.end()
#define size(x) int(x.size())

int gcd(int a, int b) { return __gcd(a, b); }
int lcm(int a, int b) { return a * (b / gcd(a, b)); }
void Yes_No(bool f) {
  if (f)
    printf("Yes\n");
  else
    printf("No\n");
}
void YES_NO(bool f) {
  if (f)
    printf("YES\n");
  else
    printf("NO\n");
}

void deb_p(pii p) { printf("pair = %d %d\n", p.fi, p.se); }
void deb_i(int i) { printf("i = %d\n", i); }
void deb_v(vector<int> v) {
  printf("v = ");
  for (auto i : v) printf("%d ", i);
  printf("\n");
}
template <typename T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool chmin(T& a, const T& b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
//-------------------ここから回答する-----------------------
vi a(3), b(3);
vvi v(3, vi(3));
bool flag = false;

void f(vi x, vi y) {
  int cta = 0;
  rep(i, 3) rep(j, 3) if (x[i] + y[j] == v[i][j]) cta++;
  rep(i, 3) {
    if (x[i] + 1 <= a[i]) {
      x[i]++;
      f(x, y);
      x[i]--;
    }
  }
  rep(i, 3) {
    if (y[i] + 1 <= b[i]) {
      y[i]++;
      f(x, y);
      y[i]--;
    }
  }
}

void solve(void) {
  rep(i, 3) rep(j, 3) cin >> v[i][j];
  rep(i, 3) a[i] = *min_element(all(v[i]));
  rep(i, 3) b[i] = min(v[0][i], min(v[1][i], v[2][i]));
  f({0, 0, 0}, {0, 0, 0});
  Yes_No(flag);
}
int main(void) { solve(); }