#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define rep(i, n) for (int i = 0; i < n; i++)
const int mo = 1e9 + 7,sz=310;
typedef long long ll;
ll n,a[sz][sz];
bool avoid[sz][sz];
ll ans=0;
void solve()
{
    bool ok=1;
    rep(i,n)rep(j,n)rep(k,n)
    {
        if(a[i][k]+a[k][j]<a[i][j])
        {
            ok=0;
            break;
        }
        if(i!=k&&j!=k)if(a[i][k]+a[k][j]==a[i][j])avoid[i][j]=1;;
    }
    rep(i,n)rep(j,n)if(!avoid[i][j])ans+=a[i][j];
    if(!ok)ans=-1;
    else ans/=2;
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(0);
    cin>>n;
    rep(i,n)rep(j,n)cin>>a[i][j];
    solve();
    cout<<ans<<endl;
    return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define rep(i, n) for (int i = 0; i < n; i++)
const int mo = 1e9 + 7,sz=310;
typedef long long ll;
ll n,a[sz][sz];
bool avoid[sz][sz];
ll ans=0;
void solve()
{
    bool ok=1;
    rep(i,n)rep(j,n)rep(k,n)
    {
        if(a[i][k]+a[k][j]<a[i][j])
        {
            ok=0;
            break;
        }
        if(i!=k&&j!=k)if(a[i][k]+a[k][j]==a[i][j])avoid[i][j]=1;;
    }
    rep(i,n)rep(j,n)if(!avoid[i][j])ans+=a[i][j];
    if(!ok)ans=-1;
    else ans/=2;
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(0);
    cin>>n;
    rep(i,n)rep(j,n)cin>>a[i][j];
    solve();
    cout<<ans<<endl;
    return 0;
}
```