#include <bits/stdc++.h>
#define ALL(A) (A).begin(),(A).end()
#define ll long long
#define rep(i,n) for(int i=0;i<(n);i++)

const ll mod = 1e9+7;
const ll INF = -1*((1LL<<63)+1) - 1e9*300-1;
const int inf = -1*((1<<31)+1);

using namespace std;

ll A[303][303];
int main(void){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	vector<vector<bool>> A_n(N,vector<bool>(N));
	rep(i,N)rep(j,N){
		cin >> A[i][j];
	}
	rep(k,N)rep(i,N)rep(j,N){
		if(A[i][j] > A[i][k] + A[k][j]){
			cout << -1 << endl;
			return 0;
		}
		if(A[i][j] == A[i][k] + A[k][j] && k!=i &&k!=j){
			A_n[i][j] = true;//辺(i,j)と(j,i)は使わなくてok
		}
	}
	ll ans = 0;
	for(int i=0; i<N; i++){
		for(int j=0; j<N; j++){
			if(A_n[i][j] == false){
				ans += A[i][j];
			}
		}
	}
	cout << ans/2LL << endl;
	
}
### Prompt

Please create a solution in Cpp 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 ALL(A) (A).begin(),(A).end()
#define ll long long
#define rep(i,n) for(int i=0;i<(n);i++)

const ll mod = 1e9+7;
const ll INF = -1*((1LL<<63)+1) - 1e9*300-1;
const int inf = -1*((1<<31)+1);

using namespace std;

ll A[303][303];
int main(void){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	vector<vector<bool>> A_n(N,vector<bool>(N));
	rep(i,N)rep(j,N){
		cin >> A[i][j];
	}
	rep(k,N)rep(i,N)rep(j,N){
		if(A[i][j] > A[i][k] + A[k][j]){
			cout << -1 << endl;
			return 0;
		}
		if(A[i][j] == A[i][k] + A[k][j] && k!=i &&k!=j){
			A_n[i][j] = true;//辺(i,j)と(j,i)は使わなくてok
		}
	}
	ll ans = 0;
	for(int i=0; i<N; i++){
		for(int j=0; j<N; j++){
			if(A_n[i][j] == false){
				ans += A[i][j];
			}
		}
	}
	cout << ans/2LL << endl;
	
}
```