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

#define INF 1.1e9
#define LINF 1.1e18
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(), (v).end()

typedef long long ll;
typedef pair<int, int> P;

ll n;
ll d[300][300];

int main(){
	cin.tie(0);
  ios::sync_with_stdio(false);

	cin>>n;
	REP(i,n) REP(j,n) cin>>d[i][j];
	ll ans=0;
	REP(i,n) {
		FOR(j,i+1,n) {
			bool flg=false;
			REP(k,n) {
				if(d[i][j]>d[i][k]+d[k][j]) {
					cout<<"-1"<<endl;
					return 0;
				}
				if(i!=k&&j!=k&&d[i][j]==d[i][k]+d[k][j]) flg=true;
			}
			if(!flg) ans+=d[i][j];
		}
	}
	cout<<ans<<endl;

	return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
#include <stdio.h>
using namespace std;

#define INF 1.1e9
#define LINF 1.1e18
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(), (v).end()

typedef long long ll;
typedef pair<int, int> P;

ll n;
ll d[300][300];

int main(){
	cin.tie(0);
  ios::sync_with_stdio(false);

	cin>>n;
	REP(i,n) REP(j,n) cin>>d[i][j];
	ll ans=0;
	REP(i,n) {
		FOR(j,i+1,n) {
			bool flg=false;
			REP(k,n) {
				if(d[i][j]>d[i][k]+d[k][j]) {
					cout<<"-1"<<endl;
					return 0;
				}
				if(i!=k&&j!=k&&d[i][j]==d[i][k]+d[k][j]) flg=true;
			}
			if(!flg) ans+=d[i][j];
		}
	}
	cout<<ans<<endl;

	return 0;
}

```