#include "bits/stdc++.h"
#define in std::cin
#define out std::cout
#define rep(i,N) for(LL i=0;i<N;++i)
typedef long long int LL;

LL N, A[312][312];
bool removeFlag[312][312];
LL ans;

bool warshall_floyd(LL V)
{
	for (LL k = 0; k < V; ++k)
	{
		for (LL i = 0; i < V; ++i)
		{
			for (LL j = 0; j < V; ++j)
			{
				if (A[i][j] == A[i][k] + A[k][j] && i != k&&k != j) { removeFlag[i][j] = true; }
				if (A[i][j] > A[i][k] + A[k][j]) { return false; }
			}
		}
	}
	return true;
}

int main()
{
	memset(removeFlag, false, sizeof(removeFlag));
	in >> N;
	rep(i, N) { rep(j, N) { in >> A[i][j]; } }

	if (!warshall_floyd(N)) { ans = -1; }
	else
	{
		rep(i, N)
		{
			rep(j, i)
			{
				if (!removeFlag[i][j]) { ans += A[i][j]; }
			}
		}
	}

	out << ans << std::endl;
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 in std::cin
#define out std::cout
#define rep(i,N) for(LL i=0;i<N;++i)
typedef long long int LL;

LL N, A[312][312];
bool removeFlag[312][312];
LL ans;

bool warshall_floyd(LL V)
{
	for (LL k = 0; k < V; ++k)
	{
		for (LL i = 0; i < V; ++i)
		{
			for (LL j = 0; j < V; ++j)
			{
				if (A[i][j] == A[i][k] + A[k][j] && i != k&&k != j) { removeFlag[i][j] = true; }
				if (A[i][j] > A[i][k] + A[k][j]) { return false; }
			}
		}
	}
	return true;
}

int main()
{
	memset(removeFlag, false, sizeof(removeFlag));
	in >> N;
	rep(i, N) { rep(j, N) { in >> A[i][j]; } }

	if (!warshall_floyd(N)) { ans = -1; }
	else
	{
		rep(i, N)
		{
			rep(j, i)
			{
				if (!removeFlag[i][j]) { ans += A[i][j]; }
			}
		}
	}

	out << ans << std::endl;
	return 0;
}
```