In Takahashi Kingdom, which once existed, there are N cities, and some pairs of cities are connected bidirectionally by roads. The following are known about the road network:

* People traveled between cities only through roads. It was possible to reach any city from any other city, via intermediate cities if necessary.
* Different roads may have had different lengths, but all the lengths were positive integers.



Snuke the archeologist found a table with N rows and N columns, A, in the ruin of Takahashi Kingdom. He thought that it represented the shortest distances between the cities along the roads in the kingdom.

Determine whether there exists a road network such that for each u and v, the integer A_{u, v} at the u-th row and v-th column of A is equal to the length of the shortest path from City u to City v. If such a network exist, find the shortest possible total length of the roads.

Constraints

* 1 \leq N \leq 300
* If i ≠ j, 1 \leq A_{i, j} = A_{j, i} \leq 10^9.
* A_{i, i} = 0

Inputs

Input is given from Standard Input in the following format:


N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
...
A_{N, 1} A_{N, 2} ... A_{N, N}


Outputs

If there exists no network that satisfies the condition, print `-1`. If it exists, print the shortest possible total length of the roads.

Examples

Input

3
0 1 3
1 0 2
3 2 0


Output

3


Input

3
0 1 3
1 0 1
3 1 0


Output

-1


Input

5
0 21 18 11 28
21 0 13 10 26
18 13 0 23 13
11 10 23 0 17
28 26 13 17 0


Output

82


Input

3
0 1000000000 1000000000
1000000000 0 1000000000
1000000000 1000000000 0


Output

3000000000
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  long long ans = 0;
  vector<vector<long long>> a(n, vector<long long>(n));
  for(int i=0;i<n;i++) for(int j=0;j<n;j++) cin >> a[i][j];
  for(int i=0;i<n;i++) {
    for(int j=i+1;j<n;j++) {
      bool need = true;
      for(int k=0;k<n;k++) {
        if(k == i || k == j) continue;
        if(a[i][j] < a[i][k] + a[k][j]) continue;
        else if(a[i][j] == a[i][k] + a[k][j]) need = false;
        else {
          ans = -1;
          break;
        }
      }
      if(ans == -1) break;
      if(need) ans += a[i][j];
    }
    if(ans == -1) break;
  }
  cout << ans << endl;
  return 0;
}
