#include <iostream>
#include <algorithm>
using namespace std;

const int N=333;
typedef long long LL;
int A[N][N],idx[N][N];

int main(){
	int n,flag=1;
	LL ans=0;
	cin>>n;
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)
	cin>>A[i][j],idx[i][j]=0;
	
	for(int k=1;k<=n;k++){
		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++){
				if(i==j||i==k||j==k) continue;
				if(A[i][k]+A[k][j]<A[i][j]) flag=0;
				if(A[i][k]+A[k][j]==A[i][j]) idx[i][j]=1;
			}
		}
	}
	
	if(!flag) {cout<<-1<<endl;return 0;}
	for(int i=1;i<=n;i++){
		for(int j=i+1;j<=n;j++){
			if(!idx[i][j]) ans+=A[i][j];
		}
	}
	
	cout<<ans<<endl;
	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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 <algorithm>
using namespace std;

const int N=333;
typedef long long LL;
int A[N][N],idx[N][N];

int main(){
	int n,flag=1;
	LL ans=0;
	cin>>n;
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)
	cin>>A[i][j],idx[i][j]=0;
	
	for(int k=1;k<=n;k++){
		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++){
				if(i==j||i==k||j==k) continue;
				if(A[i][k]+A[k][j]<A[i][j]) flag=0;
				if(A[i][k]+A[k][j]==A[i][j]) idx[i][j]=1;
			}
		}
	}
	
	if(!flag) {cout<<-1<<endl;return 0;}
	for(int i=1;i<=n;i++){
		for(int j=i+1;j<=n;j++){
			if(!idx[i][j]) ans+=A[i][j];
		}
	}
	
	cout<<ans<<endl;
	return 0;
}
```