#include <iostream>
#include <vector>
#include <cstring>
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, s, n) for(int i = s; i < n; i++)
using namespace std;

const int INF = 100000000;
int V, E;
int dist[16][16];
int dp[(1<<16)+1][16];

int rec(int bit, int v){
  if (dp[bit][v] != -1) return dp[bit][v];
  if (bit == (1<<v)){
    if (dist[0][v] == -1)
      return dp[bit][v] = INF;
    else
      return dp[bit][v] = dist[0][v];
  }
  int res = INF;
  int prev_bit = bit & ~(1<<v);
  rep(u, V){
    if (!(prev_bit & (1<<u))) continue;
    if (dist[u][v] == -1) continue;
    if (res > rec(prev_bit, u) + dist[u][v]){
      res = rec(prev_bit, u) + dist[u][v];
    }
  }
  return dp[bit][v] = res;
}

int main(){
  cin >> V >> E;
  memset(dist, -1, sizeof(dist));
  memset(dp, -1, sizeof(dp));
  int s, t, d;
  rep(i, E){
    cin >> s >> t >> d;
    dist[s][t] = d;
  }
  int res = INF;
  res = rec((1<<V)-1, 0);
  if (res == INF)
    cout << -1 << endl;
  else
    cout << res << endl;
  
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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 <vector>
#include <cstring>
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, s, n) for(int i = s; i < n; i++)
using namespace std;

const int INF = 100000000;
int V, E;
int dist[16][16];
int dp[(1<<16)+1][16];

int rec(int bit, int v){
  if (dp[bit][v] != -1) return dp[bit][v];
  if (bit == (1<<v)){
    if (dist[0][v] == -1)
      return dp[bit][v] = INF;
    else
      return dp[bit][v] = dist[0][v];
  }
  int res = INF;
  int prev_bit = bit & ~(1<<v);
  rep(u, V){
    if (!(prev_bit & (1<<u))) continue;
    if (dist[u][v] == -1) continue;
    if (res > rec(prev_bit, u) + dist[u][v]){
      res = rec(prev_bit, u) + dist[u][v];
    }
  }
  return dp[bit][v] = res;
}

int main(){
  cin >> V >> E;
  memset(dist, -1, sizeof(dist));
  memset(dp, -1, sizeof(dp));
  int s, t, d;
  rep(i, E){
    cin >> s >> t >> d;
    dist[s][t] = d;
  }
  int res = INF;
  res = rec((1<<V)-1, 0);
  if (res == INF)
    cout << -1 << endl;
  else
    cout << res << endl;
  
  return 0;
}

```