#include<bits/stdc++.h>
#define mod 1000000007
#define ll long long int
#define ar array
#define pb push_back
using namespace std;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	int n;
	cin>>n;
	ll a[n][n];
	for(int i = 0; i<n; i++)
		for(int j = 0; j<n; j++)
			cin>>a[i][j];
	int flag = 0;
	for(int k = 0; k<n; k++)
		for(int i = 0;i <n; i++)
			for(int j = 0; j<n; j++)
			{
				if(a[i][j]>a[i][k]+a[k][j])
				flag = 1;
			}
	if(flag ==1)
		cout<<-1<<endl;
	else
	{
		ll ans = 0;
		for(int i = 0; i<n; i++)
			for(int j =0; j<n; j++)
			{
				int flag2 = 0;
				for(int k =0 ; k<n; k++)
					if(a[i][j]==a[i][k]+a[k][j] && k!=i && k!=j)
						flag2 = 1;
				if(!flag2)
					ans += a[i][j];
			}

		cout<<ans/2<<endl;
	}
	return 0;
}
### Prompt

Create a solution in cpp for 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 mod 1000000007
#define ll long long int
#define ar array
#define pb push_back
using namespace std;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	int n;
	cin>>n;
	ll a[n][n];
	for(int i = 0; i<n; i++)
		for(int j = 0; j<n; j++)
			cin>>a[i][j];
	int flag = 0;
	for(int k = 0; k<n; k++)
		for(int i = 0;i <n; i++)
			for(int j = 0; j<n; j++)
			{
				if(a[i][j]>a[i][k]+a[k][j])
				flag = 1;
			}
	if(flag ==1)
		cout<<-1<<endl;
	else
	{
		ll ans = 0;
		for(int i = 0; i<n; i++)
			for(int j =0; j<n; j++)
			{
				int flag2 = 0;
				for(int k =0 ; k<n; k++)
					if(a[i][j]==a[i][k]+a[k][j] && k!=i && k!=j)
						flag2 = 1;
				if(!flag2)
					ans += a[i][j];
			}

		cout<<ans/2<<endl;
	}
	return 0;
}
```