#include<iostream>
#include<vector>
#include<utility>
#include<algorithm>
#include<set>
using namespace std ;
using ll = long long ;
using vll = vector<ll> ;
using vvll = vector<vll> ;
using pll = pair<ll,ll> ;

void chmin(ll &a,ll b){if(a>b) a = b ;}

int main(){
	ll n ; cin >> n ;
	vvll d1(n,vll(n)) ;
	for(auto &i:d1) for(auto &j:i) cin >> j ;
	vvll d2 = d1 ;
	for(int k=0;k<n;k++) for(int i=0;i<n;i++) for(int j=0;j<n;j++) chmin(d2[i][j],d2[i][k]+d2[k][j]) ;
	if(d1!=d2){
		cout << -1 << endl ;
		return 0 ;
	}
	ll ans = 0 ;
	for(int i=0;i<n;i++) for(int j=0;j<n;j++) ans += d1[i][j] ;
	for(int i=0;i<n;i++) for(int j=0;j<n;j++) for(int k=0;k<n;k++){
		if(k==i||k==j) continue ;
		if(d1[i][k]+d1[k][j]==d1[i][j]){
			ans -= d1[i][j] ;
			break ;
		}
	}
	cout << ans/2 << endl ;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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<iostream>
#include<vector>
#include<utility>
#include<algorithm>
#include<set>
using namespace std ;
using ll = long long ;
using vll = vector<ll> ;
using vvll = vector<vll> ;
using pll = pair<ll,ll> ;

void chmin(ll &a,ll b){if(a>b) a = b ;}

int main(){
	ll n ; cin >> n ;
	vvll d1(n,vll(n)) ;
	for(auto &i:d1) for(auto &j:i) cin >> j ;
	vvll d2 = d1 ;
	for(int k=0;k<n;k++) for(int i=0;i<n;i++) for(int j=0;j<n;j++) chmin(d2[i][j],d2[i][k]+d2[k][j]) ;
	if(d1!=d2){
		cout << -1 << endl ;
		return 0 ;
	}
	ll ans = 0 ;
	for(int i=0;i<n;i++) for(int j=0;j<n;j++) ans += d1[i][j] ;
	for(int i=0;i<n;i++) for(int j=0;j<n;j++) for(int k=0;k<n;k++){
		if(k==i||k==j) continue ;
		if(d1[i][k]+d1[k][j]==d1[i][j]){
			ans -= d1[i][j] ;
			break ;
		}
	}
	cout << ans/2 << endl ;
}
```