#include<bits/stdc++.h>

using namespace std;
using ll = long long;
using P = pair<int,int>;
using PL = pair<ll,ll>;
const ll INF = 1e15;


int main()
{
	int n;
	cin >> n;
	vector<vector<ll>> dist(n, vector<ll>(n));
	ll sum =0;
	for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) {
		cin >> dist[i][j];
		sum += dist[i][j];
	}
	set<P> edge;

	for(int k = 0; k < n; k++) {
		for(int i = 0; i < n; i++) {
			for(int j = 0; j < n; j++) {

				if(dist[i][j] > dist[i][k] + dist[k][j]) {
					cout << -1 << endl;
					return 0;
				}

				if(dist[i][j] == dist[i][k] + dist[k][j] && dist[i][k] > 0 && dist[k][j] > 0) {
					edge.emplace(i, j);
				}

			}
		}
	}
	for(auto i : edge) {
		sum -= dist[i.first][i.second];
	}
	cout << sum / 2 << endl;

}
### Prompt

In Cpp, your task is to solve 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>

using namespace std;
using ll = long long;
using P = pair<int,int>;
using PL = pair<ll,ll>;
const ll INF = 1e15;


int main()
{
	int n;
	cin >> n;
	vector<vector<ll>> dist(n, vector<ll>(n));
	ll sum =0;
	for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) {
		cin >> dist[i][j];
		sum += dist[i][j];
	}
	set<P> edge;

	for(int k = 0; k < n; k++) {
		for(int i = 0; i < n; i++) {
			for(int j = 0; j < n; j++) {

				if(dist[i][j] > dist[i][k] + dist[k][j]) {
					cout << -1 << endl;
					return 0;
				}

				if(dist[i][j] == dist[i][k] + dist[k][j] && dist[i][k] > 0 && dist[k][j] > 0) {
					edge.emplace(i, j);
				}

			}
		}
	}
	for(auto i : edge) {
		sum -= dist[i.first][i.second];
	}
	cout << sum / 2 << endl;

}
```