#define REP(i,n) for (int i=0;i<(n);i++)
#include <bits/stdc++.h>
using namespace std;
int tmp,n,a[300][300],wf[300][300];

int main(){
	scanf("%d", &n);
	long long ret = 0;
	REP(i,n){ REP(j,n){
		scanf("%d", &tmp);
		a[i][j] = tmp;
		wf[i][j] = tmp;
	} }

	REP(k,n){ REP(i,n){ REP(j,n){
		if(wf[i][j] > wf[i][k] + wf[k][j]){
			wf[i][j] = wf[i][k] + wf[k][j];
		}
	} } }

	REP(i,n){ REP(j,n){ REP(k,n){
		if (wf[i][j] < a[i][j]) { printf("%s","-1\n"); return 0;}
		if ( i!=k & j!=k ){
			if (wf[i][j] == wf[i][k] + wf[k][j]) {break;}
		}
		if (k==n-1){ret += wf[i][j];}
	} } }
	ret /= 2;
	printf("%lld\n",ret);
}

### Prompt

Generate 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
#define REP(i,n) for (int i=0;i<(n);i++)
#include <bits/stdc++.h>
using namespace std;
int tmp,n,a[300][300],wf[300][300];

int main(){
	scanf("%d", &n);
	long long ret = 0;
	REP(i,n){ REP(j,n){
		scanf("%d", &tmp);
		a[i][j] = tmp;
		wf[i][j] = tmp;
	} }

	REP(k,n){ REP(i,n){ REP(j,n){
		if(wf[i][j] > wf[i][k] + wf[k][j]){
			wf[i][j] = wf[i][k] + wf[k][j];
		}
	} } }

	REP(i,n){ REP(j,n){ REP(k,n){
		if (wf[i][j] < a[i][j]) { printf("%s","-1\n"); return 0;}
		if ( i!=k & j!=k ){
			if (wf[i][j] == wf[i][k] + wf[k][j]) {break;}
		}
		if (k==n-1){ret += wf[i][j];}
	} } }
	ret /= 2;
	printf("%lld\n",ret);
}

```