#include <iostream>
#include <cstdio>
using namespace std;
int main() {
	long N;
	long *A;
	long result=0;
	bool isMin=true;
	cin >> N;
	A = new long[N*N];

	for (size_t i = 0; i < N; i++) {
		for (size_t j = 0; j < N; j++) {
			cin >> A[i*N+j];
		}
	}

	for (size_t i = 0; i < N-1; i++) {
		for (size_t j = i+1; j < N; j++) {
			for (size_t k = 0; k < N; k++) {
				if (k==i || k==j) {
					continue;
				}
				if (A[i*N+j]>A[i*N+k]+A[k*N+j]) {
					result=-1;
					i=N-1;
					j=i;
					k=N;
					continue;
				}else if(A[i*N+j]==A[i*N+k]+A[k*N+j]){
					// 最小値が大体できるか
					isMin=false;
				}
			}
			if (isMin) {
				result+=A[i*N+j];
			}else{
				isMin=true;
			}

		}
	}

	cout << result << endl;
	return 0;
}

### Prompt

Your challenge is to write 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
#include <iostream>
#include <cstdio>
using namespace std;
int main() {
	long N;
	long *A;
	long result=0;
	bool isMin=true;
	cin >> N;
	A = new long[N*N];

	for (size_t i = 0; i < N; i++) {
		for (size_t j = 0; j < N; j++) {
			cin >> A[i*N+j];
		}
	}

	for (size_t i = 0; i < N-1; i++) {
		for (size_t j = i+1; j < N; j++) {
			for (size_t k = 0; k < N; k++) {
				if (k==i || k==j) {
					continue;
				}
				if (A[i*N+j]>A[i*N+k]+A[k*N+j]) {
					result=-1;
					i=N-1;
					j=i;
					k=N;
					continue;
				}else if(A[i*N+j]==A[i*N+k]+A[k*N+j]){
					// 最小値が大体できるか
					isMin=false;
				}
			}
			if (isMin) {
				result+=A[i*N+j];
			}else{
				isMin=true;
			}

		}
	}

	cout << result << endl;
	return 0;
}

```