#include<bits/stdc++.h>
using namespace std;
using ll = int64_t;

int main(int , char **)
{
	int N;
	cin >> N;
	ll A[300][300]={},B[300][300]={};

	for(int y=0;y<N;y++){
		for(int x=0;x<N;x++){
			int a;
			cin >> a;
			A[y][x]=a;
			B[y][x]=a;
		}
	}

	for(int k=0;k<N;k++){
		for(int y=0;y<N;y++){
			for(int x=0;x<N;x++){
				if ( A[y][k] + A[k][x] < A[y][x] ){printf("-1\n");return 0;}
				if ( A[y][k] + A[k][x] == A[y][x]){
					if (( B[y][k]!=0 )&&( B[k][x]!=0 ))	{
						B[y][x] = 0;
					}
				}
			}
		}
	}

	ll ans = 0;
	for(int y=0;y<N;y++){
		for(int x=0;x<y;x++){
			ans += B[y][x];
		}
	}

	printf("%ld\n",ans);
	return 0;
}

### Prompt

Please create a solution in cpp 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>
using namespace std;
using ll = int64_t;

int main(int , char **)
{
	int N;
	cin >> N;
	ll A[300][300]={},B[300][300]={};

	for(int y=0;y<N;y++){
		for(int x=0;x<N;x++){
			int a;
			cin >> a;
			A[y][x]=a;
			B[y][x]=a;
		}
	}

	for(int k=0;k<N;k++){
		for(int y=0;y<N;y++){
			for(int x=0;x<N;x++){
				if ( A[y][k] + A[k][x] < A[y][x] ){printf("-1\n");return 0;}
				if ( A[y][k] + A[k][x] == A[y][x]){
					if (( B[y][k]!=0 )&&( B[k][x]!=0 ))	{
						B[y][x] = 0;
					}
				}
			}
		}
	}

	ll ans = 0;
	for(int y=0;y<N;y++){
		for(int x=0;x<y;x++){
			ans += B[y][x];
		}
	}

	printf("%ld\n",ans);
	return 0;
}

```