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

const int MAX_V = 18;
const int INF = 1e18;

int V,E;
int d[MAX_V][MAX_V];
int dp[1 << MAX_V][MAX_V];

int rec(int S, int v){
    if(dp[S][v] >= 0){
        return dp[S][v];
    }

    if(S == (1 << V) - 1 && v == 0){
        return dp[S][v] = 0;
    }

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

signed main() {
    cin >> V >> E;
    memset(dp,-1,sizeof(dp));

    rep(i,MAX_V) rep(j,MAX_V) d[i][j] = INF;
    for (int i = 0; i < E; i++) {
        int a,b,c;
        cin >> a >> b >> c;
        d[a][b] = c;
    }
    int ans = rec(0,0);
    if(ans == INF) ans = -1;
    cout << ans << endl;
}

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

const int MAX_V = 18;
const int INF = 1e18;

int V,E;
int d[MAX_V][MAX_V];
int dp[1 << MAX_V][MAX_V];

int rec(int S, int v){
    if(dp[S][v] >= 0){
        return dp[S][v];
    }

    if(S == (1 << V) - 1 && v == 0){
        return dp[S][v] = 0;
    }

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

signed main() {
    cin >> V >> E;
    memset(dp,-1,sizeof(dp));

    rep(i,MAX_V) rep(j,MAX_V) d[i][j] = INF;
    for (int i = 0; i < E; i++) {
        int a,b,c;
        cin >> a >> b >> c;
        d[a][b] = c;
    }
    int ans = rec(0,0);
    if(ans == INF) ans = -1;
    cout << ans << endl;
}

```