#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;

#define MOD 1000000007
#define REP(i, N) for (int i = 0; i < N; ++i)
#define REP1(i, N) for (int i = 1; i <= N; ++i)
#define RREP(i, N) for (int i = N - 1; i >= 0; --i)
#define ALL(a) a.begin(), a.end()

const int INF = 1e9;
int V, E;
int dist[15][15];
int dp[1 << 15][15];

int rec(int S, int v) {
  if (0 <= dp[S][v]) {
    return dp[S][v];
  }
  if (S == (1 << V) - 1 && v == 0) {
    return dp[S][v] = 0;
  }
  int res = INF;
  REP(u, V) {
    if (!(S >> u & 1)) {
      res = min(res, rec(S | 1 << u, u) + dist[v][u]);
    }
  }
  return dp[S][v] = res;
}

int main() {
  cin >> V >> E;
  REP(i, V) REP(j, V) {
    dist[i][j] = INF;
    if (i == j) dist[i][j] = 0;
  }
  REP(i, E) {
    int s, t, d;
    cin >> s >> t >> d;
    dist[s][t] = d;
  }
  REP(i, 1 << V) REP(j, V) dp[i][j] = -1;
  int ans = rec(0, 0);
  cout << (ans == INF ? -1 : ans) << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 <algorithm>
#include <iostream>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;

#define MOD 1000000007
#define REP(i, N) for (int i = 0; i < N; ++i)
#define REP1(i, N) for (int i = 1; i <= N; ++i)
#define RREP(i, N) for (int i = N - 1; i >= 0; --i)
#define ALL(a) a.begin(), a.end()

const int INF = 1e9;
int V, E;
int dist[15][15];
int dp[1 << 15][15];

int rec(int S, int v) {
  if (0 <= dp[S][v]) {
    return dp[S][v];
  }
  if (S == (1 << V) - 1 && v == 0) {
    return dp[S][v] = 0;
  }
  int res = INF;
  REP(u, V) {
    if (!(S >> u & 1)) {
      res = min(res, rec(S | 1 << u, u) + dist[v][u]);
    }
  }
  return dp[S][v] = res;
}

int main() {
  cin >> V >> E;
  REP(i, V) REP(j, V) {
    dist[i][j] = INF;
    if (i == j) dist[i][j] = 0;
  }
  REP(i, E) {
    int s, t, d;
    cin >> s >> t >> d;
    dist[s][t] = d;
  }
  REP(i, 1 << V) REP(j, V) dp[i][j] = -1;
  int ans = rec(0, 0);
  cout << (ans == INF ? -1 : ans) << endl;
  return 0;
}

```