#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

const int inf = 987654321;
int n, m;
vector< vector<int> > G;
vector< vector<int> > dp;

int rec(int visited, int now) {
  if (dp[visited][now] != -1) return dp[visited][now];
  if (visited == (1<<n)-1 && now == 0) return dp[visited][now] = 0;
  int res = inf;
  for (int i = 0; i < n; i++) {
    if (visited >> i & 1) continue; /* ??¢????¨??????? */
    res = min(res, rec(visited | (1 << i), i) + G[now][i]);
  }
  return dp[visited][now] = res;
}

int main(void) {
  cin >> n >> m;
  G.assign(n, vector<int>(n, inf));
  for (int i = 0; i < m; i++) {
    int p, q, c;
    cin >> p >> q >> c;
    G[p][q] = c;
  }
  dp.assign(1 << n, vector<int>(n, -1));
  int res = rec(0, 0);
  if(res != inf) cout << res << endl;
  else cout << -1 << endl;
  return 0;
}
### Prompt

In cpp, your task is to solve the following problem:

For a given weighted directed graph G(V, E), find the distance of the shortest route that meets the following criteria:

* It is a closed cycle where it ends at the same point it starts.
* It visits each vertex exactly once.

Constraints

* 2 ≤ |V| ≤ 15
* 0 ≤ di ≤ 1,000
* There are no multiedge

Input


|V| |E|
s0 t0 d0
s1 t1 d1
:
s|E|-1 t|E|-1 d|E|-1


|V| is the number of vertices and |E| is the number of edges in the graph. The graph vertices are named with the numbers 0, 1,..., |V|-1 respectively.

si and ti represent source and target vertices of i-th edge (directed) and di represents the distance between si and ti (the i-th edge).

Output

Print the shortest distance in a line. If there is no solution, print -1.

Examples

Input

4 6
0 1 2
1 2 3
1 3 9
2 0 1
2 3 6
3 2 4


Output

16


Input

3 3
0 1 1
1 2 1
0 2 1


Output

-1

### Response

```cpp
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

const int inf = 987654321;
int n, m;
vector< vector<int> > G;
vector< vector<int> > dp;

int rec(int visited, int now) {
  if (dp[visited][now] != -1) return dp[visited][now];
  if (visited == (1<<n)-1 && now == 0) return dp[visited][now] = 0;
  int res = inf;
  for (int i = 0; i < n; i++) {
    if (visited >> i & 1) continue; /* ??¢????¨??????? */
    res = min(res, rec(visited | (1 << i), i) + G[now][i]);
  }
  return dp[visited][now] = res;
}

int main(void) {
  cin >> n >> m;
  G.assign(n, vector<int>(n, inf));
  for (int i = 0; i < m; i++) {
    int p, q, c;
    cin >> p >> q >> c;
    G[p][q] = c;
  }
  dp.assign(1 << n, vector<int>(n, -1));
  int res = rec(0, 0);
  if(res != inf) cout << res << endl;
  else cout << -1 << endl;
  return 0;
}
```