#include <iostream>
using namespace std;

int n;
long long g[300][300];
bool calc[300][300];

int main() {
    cin.tie(0); ios_base::sync_with_stdio(false);
    cin >> n;
    for(int i=0; i < n; ++i) 
        for(int j=0; j < n; ++j) cin >> g[i][j];

    for(int k = 0; k < n; ++k)
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < n; ++j) {
                if(i == j || j == k || k == i) continue;
                if(g[i][j] > g[i][k] + g[k][j]) {cout << -1 << '\n'; return 0;}
                if(g[i][j] == g[i][k] + g[k][j]) calc[i][j] = true;
            }

    long long ans = 0;
    for(int i=0; i < n; ++i) 
        for(int j=0; j < i; ++j) 
            if(!calc[i][j]) ans += g[i][j];
    cout << ans << '\n';
    return 0;
}
### Prompt

Your challenge is to write 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 <iostream>
using namespace std;

int n;
long long g[300][300];
bool calc[300][300];

int main() {
    cin.tie(0); ios_base::sync_with_stdio(false);
    cin >> n;
    for(int i=0; i < n; ++i) 
        for(int j=0; j < n; ++j) cin >> g[i][j];

    for(int k = 0; k < n; ++k)
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < n; ++j) {
                if(i == j || j == k || k == i) continue;
                if(g[i][j] > g[i][k] + g[k][j]) {cout << -1 << '\n'; return 0;}
                if(g[i][j] == g[i][k] + g[k][j]) calc[i][j] = true;
            }

    long long ans = 0;
    for(int i=0; i < n; ++i) 
        for(int j=0; j < i; ++j) 
            if(!calc[i][j]) ans += g[i][j];
    cout << ans << '\n';
    return 0;
}
```