#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define irep(i, n) for (int i = (n); i >= 0; i--)
using namespace std;
using ll = long long;
using P = pair<int, int>;
const int INF = 1 << 25;
int n, a[350][350]; 
bool road[350][350];
int main() {
  cin >> n;
  rep(i,n) rep(j,n) cin >> a[i][j];
  rep(i,n) rep(j,n) if (i!=j) road[i][j] = true;
  
  rep(i,n) {
    rep(j,n) {
      rep(k,n) {
        if (a[i][j] > a[i][k]+a[k][j]) {
          cout << -1 << endl;
          return 0;
        }
        if (a[i][j] == a[i][k]+a[k][j] && k!=i && k!=j) road[i][j] = false;
      }
    }
  }
  ll len = 0;
  rep(i,n) rep(j,n) {
    if (road[i][j]) len += a[i][j];
  }
  len /= 2;
  cout << len << endl;
  return 0;
}
### Prompt

Please create a solution in Cpp to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define irep(i, n) for (int i = (n); i >= 0; i--)
using namespace std;
using ll = long long;
using P = pair<int, int>;
const int INF = 1 << 25;
int n, a[350][350]; 
bool road[350][350];
int main() {
  cin >> n;
  rep(i,n) rep(j,n) cin >> a[i][j];
  rep(i,n) rep(j,n) if (i!=j) road[i][j] = true;
  
  rep(i,n) {
    rep(j,n) {
      rep(k,n) {
        if (a[i][j] > a[i][k]+a[k][j]) {
          cout << -1 << endl;
          return 0;
        }
        if (a[i][j] == a[i][k]+a[k][j] && k!=i && k!=j) road[i][j] = false;
      }
    }
  }
  ll len = 0;
  rep(i,n) rep(j,n) {
    if (road[i][j]) len += a[i][j];
  }
  len /= 2;
  cout << len << endl;
  return 0;
}
```