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

#define N 300

int n;
int a[N][N];

int main()
{
	cin >> n;
	for( int i = 0; i < n; i++ ) {
		for( int j = 0; j < n; j++ ) {
			cin >> a[i][j];
		}
	}

	int flag = 0;
	long long ans = 0;
	for( int i = 0; i < n; i++ ) {
		for( int j = i + 1; j < n; j++ ) {
			if( i == j ) {
				continue;
			}
			int flag2 = 0;
			for( int k = 0; k < n; k++ ) {
				if( k == i || k == j ) {
					continue;
				}
				if( a[i][j] > a[i][k] + a[k][j] ) {
					flag = 1;
					break;
				}
				if( a[i][j] == a[i][k] + a[k][j] ) {
					flag2 = 1;
				}
			}
			if( flag ) {
				break;
			}
			if( flag2 == 0 ) {
				ans += a[i][j];
			}
		}
		if( flag ) {
			break;
		}
	}
	if( flag ) {
		ans = -1;
	}

	cout << ans << endl;

	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;

#define N 300

int n;
int a[N][N];

int main()
{
	cin >> n;
	for( int i = 0; i < n; i++ ) {
		for( int j = 0; j < n; j++ ) {
			cin >> a[i][j];
		}
	}

	int flag = 0;
	long long ans = 0;
	for( int i = 0; i < n; i++ ) {
		for( int j = i + 1; j < n; j++ ) {
			if( i == j ) {
				continue;
			}
			int flag2 = 0;
			for( int k = 0; k < n; k++ ) {
				if( k == i || k == j ) {
					continue;
				}
				if( a[i][j] > a[i][k] + a[k][j] ) {
					flag = 1;
					break;
				}
				if( a[i][j] == a[i][k] + a[k][j] ) {
					flag2 = 1;
				}
			}
			if( flag ) {
				break;
			}
			if( flag2 == 0 ) {
				ans += a[i][j];
			}
		}
		if( flag ) {
			break;
		}
	}
	if( flag ) {
		ans = -1;
	}

	cout << ans << endl;

	return 0;
}

```