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

int main() {
  int n; cin>>n;
  vector<vector<ll>> A(n, vector<ll>(n));
  rep(i, n) rep(j, n) cin>>A[i][j];

  ll INF=1e15;
  bool ok=true;
  vector<vector<ll>> dp(n, vector<ll>(n, INF));
  rep(i, n) rep(j, n) dp[i][j]=A[i][j];

  rep(k, n) {
    rep(i, n) {
      rep(j, n) {
        if(k==i||k==j||i==j) continue;
        if (dp[i][k]==INF||dp[k][j]==INF) continue;
        if (A[i][j]>dp[i][k]+dp[k][j]) ok=false;
        if (A[i][j]==dp[i][k]+dp[k][j]) dp[i][j]=INF;
      }
    }
  }
  if (ok) {
    ll res=0;
    rep(i, n) {
      rep(j, n) {
        if (dp[i][j]!=INF) res+=dp[i][j];
      }
    }
    cout<<res/2<<endl;
  }
  else cout<<-1<<endl;
}

### Prompt

Please formulate a cpp solution 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 ll long long
using namespace std;

int main() {
  int n; cin>>n;
  vector<vector<ll>> A(n, vector<ll>(n));
  rep(i, n) rep(j, n) cin>>A[i][j];

  ll INF=1e15;
  bool ok=true;
  vector<vector<ll>> dp(n, vector<ll>(n, INF));
  rep(i, n) rep(j, n) dp[i][j]=A[i][j];

  rep(k, n) {
    rep(i, n) {
      rep(j, n) {
        if(k==i||k==j||i==j) continue;
        if (dp[i][k]==INF||dp[k][j]==INF) continue;
        if (A[i][j]>dp[i][k]+dp[k][j]) ok=false;
        if (A[i][j]==dp[i][k]+dp[k][j]) dp[i][j]=INF;
      }
    }
  }
  if (ok) {
    ll res=0;
    rep(i, n) {
      rep(j, n) {
        if (dp[i][j]!=INF) res+=dp[i][j];
      }
    }
    cout<<res/2<<endl;
  }
  else cout<<-1<<endl;
}

```