#include <bits/stdc++.h>
using namespace std;
#define int long long
const int inf = 1e18;
#define rep(i,n) for(int i = 0;i < n;i++)

int a[300][300], G[300][300];
signed main(){
	int n;
	scanf("%lld", &n);
	rep(i,n)rep(j,n)scanf("%lld", &a[i][j]);
	rep(i,n)rep(j,n)G[i][j] = a[i][j];
	rep(i,n)rep(u,n)rep(t,n){
		G[u][t] = min(G[u][t], G[u][i]+G[i][t]);
	}
	/*rep(i,n){
		rep(j,n)printf("%d ", G[i][j]);
		printf("\n");
	}*/
	int f = 1;
	rep(i,n)rep(j,n)if(a[i][j] != G[i][j])f = 0;
	if(f == 0){
		printf("-1\n");
		return 0;
	}
	int ans = 0;
	rep(u,n)rep(t,n){
		if(u >= t)continue;
		int mn = inf;
		rep(i,n){
			if(u == i || i == t)continue;
			mn = min(mn, G[u][i]+G[i][t]);
		}
		if(G[u][t] < mn)ans += G[u][t];
	}
	printf("%lld\n", ans);
	return 0;
}
### 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>
using namespace std;
#define int long long
const int inf = 1e18;
#define rep(i,n) for(int i = 0;i < n;i++)

int a[300][300], G[300][300];
signed main(){
	int n;
	scanf("%lld", &n);
	rep(i,n)rep(j,n)scanf("%lld", &a[i][j]);
	rep(i,n)rep(j,n)G[i][j] = a[i][j];
	rep(i,n)rep(u,n)rep(t,n){
		G[u][t] = min(G[u][t], G[u][i]+G[i][t]);
	}
	/*rep(i,n){
		rep(j,n)printf("%d ", G[i][j]);
		printf("\n");
	}*/
	int f = 1;
	rep(i,n)rep(j,n)if(a[i][j] != G[i][j])f = 0;
	if(f == 0){
		printf("-1\n");
		return 0;
	}
	int ans = 0;
	rep(u,n)rep(t,n){
		if(u >= t)continue;
		int mn = inf;
		rep(i,n){
			if(u == i || i == t)continue;
			mn = min(mn, G[u][i]+G[i][t]);
		}
		if(G[u][t] < mn)ans += G[u][t];
	}
	printf("%lld\n", ans);
	return 0;
}
```