#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (n); i++)

int main() {
    int n;
    cin >> n;
    vector<vector<ll>> adj(n, vector<ll>(n));
    vector<vector<ll>> mat(n, vector<ll>(n, -1));
    rep(i, n) rep(j, n) cin >> adj[i][j];

    bool shortest = true;
    rep(k, n) rep(i, n) rep(j, n) {
        if(adj[i][j] > adj[i][k] + adj[k][j]){
            shortest = false;
            break;
        }
        else if(k!=i && k!=j&&adj[i][j] == adj[i][k] + adj[k][j]){
            mat[i][j] = k;
        }
    }
    if(!shortest){
        puts("-1");
        return 0;
    }
    ll dist = 0;
    rep(i,n) rep(j,n){
        if (mat[i][j] != -1) continue;
        dist += adj[i][j];
    }
    cout << dist / 2 << endl;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (n); i++)

int main() {
    int n;
    cin >> n;
    vector<vector<ll>> adj(n, vector<ll>(n));
    vector<vector<ll>> mat(n, vector<ll>(n, -1));
    rep(i, n) rep(j, n) cin >> adj[i][j];

    bool shortest = true;
    rep(k, n) rep(i, n) rep(j, n) {
        if(adj[i][j] > adj[i][k] + adj[k][j]){
            shortest = false;
            break;
        }
        else if(k!=i && k!=j&&adj[i][j] == adj[i][k] + adj[k][j]){
            mat[i][j] = k;
        }
    }
    if(!shortest){
        puts("-1");
        return 0;
    }
    ll dist = 0;
    rep(i,n) rep(j,n){
        if (mat[i][j] != -1) continue;
        dist += adj[i][j];
    }
    cout << dist / 2 << endl;
}
```