#include<cstdio>
#include<queue>

using namespace std;

int d[310][310];
int v[310][310];
queue<int>q;

int main()
{
	int n;
	scanf("%d",&n);
	bool flag=0;
	for(int i=1;i<=n;++i)
	for(int j=1;j<=n;++j)
	scanf("%d",&d[i][j]);
	for(int k=1;k<=n;++k)
	for(int i=1;i<=n;++i)
	for(int j=1;j<=n;++j)
	{
		if(d[i][j]>d[i][k]+d[k][j])return !printf("-1");
		if((d[i][j]==d[i][k]+d[k][j])&&d[i][k]!=0&&d[i][j]!=0&&d[k][j]!=0)v[i][j]=v[j][i]=1;
	}
	long long ans=0;
	q.push(1);
	while(!q.empty())
	{
		int u=q.front();q.pop();
		for(int i=1;i<=n;++i)
		{
			if(v[u][i]||v[i][u])continue;
			if(d[u][i])
			{
				ans+=d[u][i];
				q.push(i);
				v[u][i]=v[i][u]=1;
			}
		}
	}
	printf("%lld",ans);
	return 0;
}
### Prompt

Your challenge is to write 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<cstdio>
#include<queue>

using namespace std;

int d[310][310];
int v[310][310];
queue<int>q;

int main()
{
	int n;
	scanf("%d",&n);
	bool flag=0;
	for(int i=1;i<=n;++i)
	for(int j=1;j<=n;++j)
	scanf("%d",&d[i][j]);
	for(int k=1;k<=n;++k)
	for(int i=1;i<=n;++i)
	for(int j=1;j<=n;++j)
	{
		if(d[i][j]>d[i][k]+d[k][j])return !printf("-1");
		if((d[i][j]==d[i][k]+d[k][j])&&d[i][k]!=0&&d[i][j]!=0&&d[k][j]!=0)v[i][j]=v[j][i]=1;
	}
	long long ans=0;
	q.push(1);
	while(!q.empty())
	{
		int u=q.front();q.pop();
		for(int i=1;i<=n;++i)
		{
			if(v[u][i]||v[i][u])continue;
			if(d[u][i])
			{
				ans+=d[u][i];
				q.push(i);
				v[u][i]=v[i][u]=1;
			}
		}
	}
	printf("%lld",ans);
	return 0;
}
```