#include <bits/stdc++.h>

#define F first
#define S second

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;

const int N = 310, M = 110;

int n;
int g[N][N];
bool st[N][N];

int main()
{
	cin >> n;
	for (int i = 1; i <= n; i ++ )
		for (int j = 1; j <= n; j ++ )
			cin >> g[i][j];
	
	for (int i = 1; i <= n; i ++ )
		for (int j = 1; j <= n; j ++ )
			for (int k = 1; k <= n; k ++ )
				if (k != j && k != i)
				{
					if (g[i][k] + g[k][j] < g[i][j])
					{
						puts("-1");
						return 0;
					}
					else if (g[i][k] + g[k][j] == g[i][j]) st[i][j] = true;
				}
				
	LL res = 0;
	for (int i = 1; i <= n; i ++ )
		for (int j = i + 1; j <= n; j ++ )
			if (!st[i][j])
				res += g[i][j];
	
	cout << res << endl;
	return 0;
}
### Prompt

Your task is to create 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>

#define F first
#define S second

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;

const int N = 310, M = 110;

int n;
int g[N][N];
bool st[N][N];

int main()
{
	cin >> n;
	for (int i = 1; i <= n; i ++ )
		for (int j = 1; j <= n; j ++ )
			cin >> g[i][j];
	
	for (int i = 1; i <= n; i ++ )
		for (int j = 1; j <= n; j ++ )
			for (int k = 1; k <= n; k ++ )
				if (k != j && k != i)
				{
					if (g[i][k] + g[k][j] < g[i][j])
					{
						puts("-1");
						return 0;
					}
					else if (g[i][k] + g[k][j] == g[i][j]) st[i][j] = true;
				}
				
	LL res = 0;
	for (int i = 1; i <= n; i ++ )
		for (int j = i + 1; j <= n; j ++ )
			if (!st[i][j])
				res += g[i][j];
	
	cout << res << endl;
	return 0;
}
```