#include <cstdio>
#include <vector>
#include <algorithm>
#define maxn 305
using namespace std;

int a[maxn][maxn], n;
vector<pair<int, pair<int, int> > > distances;

int main(int argc, char** argv)
{
	scanf("%d", &n);
	for(int i=1; i<=n; i++)
		for(int j=1; j<=n; j++)
		{
			scanf("%d", a[i] + j);
			if(i < j) distances.emplace_back(a[i][j], make_pair(i, j));
		}
	sort(distances.begin(), distances.end());
	long long ans = 0;
	for(auto i: distances)
	{
		int x = i.second.first, y = i.second.second;
		bool add = true;
		for(int j=1; j<=n; j++)
			if(j != x && j != y)
			{
				if(a[x][j] + a[j][y] < a[x][y])
				{
					puts("-1");
					return 0;
				}
				if(a[x][j] + a[j][y] == a[x][y]) add = false;
			}
		if(add) ans += a[x][y];
	}
	printf("%lld\n", ans);
	return 0;
}
### Prompt

Please provide a cpp coded solution 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 <cstdio>
#include <vector>
#include <algorithm>
#define maxn 305
using namespace std;

int a[maxn][maxn], n;
vector<pair<int, pair<int, int> > > distances;

int main(int argc, char** argv)
{
	scanf("%d", &n);
	for(int i=1; i<=n; i++)
		for(int j=1; j<=n; j++)
		{
			scanf("%d", a[i] + j);
			if(i < j) distances.emplace_back(a[i][j], make_pair(i, j));
		}
	sort(distances.begin(), distances.end());
	long long ans = 0;
	for(auto i: distances)
	{
		int x = i.second.first, y = i.second.second;
		bool add = true;
		for(int j=1; j<=n; j++)
			if(j != x && j != y)
			{
				if(a[x][j] + a[j][y] < a[x][y])
				{
					puts("-1");
					return 0;
				}
				if(a[x][j] + a[j][y] == a[x][y]) add = false;
			}
		if(add) ans += a[x][y];
	}
	printf("%lld\n", ans);
	return 0;
}
```