#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;

const int MAX_N = 15;
int dist[MAX_N][MAX_N];  // dist[i][j] i から j への距離

// dp[i][j]: 状態iでjにいるときの最短経路
// 状態iはビットで管理: 00011->点0と点1は通貨済
int dp[1 << MAX_N][MAX_N];
int n, m;
const int inf = 1001001;

// 巡回セールスマン問題
int dfs(int s, int v) {
  if (dp[s][v] >= 0) return dp[s][v];
  if (s == (1 << n) - 1 && v == 0) return dp[s][v] = 0;

  int ans = inf;
  rep(u, n) {
    // uに行ったことがない場合
    if (!(s >> u & 1)) ans = min(ans, dfs(s | 1 << u, u) + dist[v][u]);
  }

  return dp[s][v] = ans;
}

int main() {
  cin >> n >> m;
  memset(dp, -1, sizeof(dp));
  // パスがない場合は距離がinfになる
  fill(dist[0], dist[0] + MAX_N * MAX_N, inf);

  rep(i, m) {
    int f, t, c;
    cin >> f >> t >> c;
    dist[f][t] = c;
  }
  int ans = dfs(0, 0);
  ans = ans == inf ? -1 : ans;
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;

const int MAX_N = 15;
int dist[MAX_N][MAX_N];  // dist[i][j] i から j への距離

// dp[i][j]: 状態iでjにいるときの最短経路
// 状態iはビットで管理: 00011->点0と点1は通貨済
int dp[1 << MAX_N][MAX_N];
int n, m;
const int inf = 1001001;

// 巡回セールスマン問題
int dfs(int s, int v) {
  if (dp[s][v] >= 0) return dp[s][v];
  if (s == (1 << n) - 1 && v == 0) return dp[s][v] = 0;

  int ans = inf;
  rep(u, n) {
    // uに行ったことがない場合
    if (!(s >> u & 1)) ans = min(ans, dfs(s | 1 << u, u) + dist[v][u]);
  }

  return dp[s][v] = ans;
}

int main() {
  cin >> n >> m;
  memset(dp, -1, sizeof(dp));
  // パスがない場合は距離がinfになる
  fill(dist[0], dist[0] + MAX_N * MAX_N, inf);

  rep(i, m) {
    int f, t, c;
    cin >> f >> t >> c;
    dist[f][t] = c;
  }
  int ans = dfs(0, 0);
  ans = ans == inf ? -1 : ans;
  cout << ans << endl;
  return 0;
}

```