#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

vector<vector<ll>> dp((1 << 15),vector<ll>(15,-1));
vector<vector<pair<ll, ll>>> graph(15);
ll n;

ll solve(ll bit, ll v)
{
    if(dp.at(bit).at(v) != -1){
        return dp.at(bit).at(v);
    }
    if(bit == (1 << n) -1 && v == 0){
        return dp.at(bit).at(v) = 0;
    }
    ll ans = 1e9+7;
    ll i,j;
    for(i = 0;i < graph.at(v).size();++i){
        ll s = graph.at(v).at(i).first;
        if(bit & (1 << s)) continue;
        ll d = graph.at(v).at(i).second;
        ans = min(ans, solve(bit | 1 << s, s) + d);
    }
    return dp.at(bit).at(v) = ans;
}

int main()
{
    ll i,j;
    ll e;
    cin >> n >> e;
    for(i = 0;i < e;++i){
        ll s,t,d;
        cin >> s >> t >> d;
        graph.at(s).emplace_back(t, d);
    }
    ll ans = min((ll)1e9+7, solve(0LL, 0LL));
    if(ans == 1e9+7){
        cout << -1 << endl;
    }else{
        cout << ans << endl;
    }

    return 0;
}

### Prompt

Create a solution in CPP for 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;

typedef long long ll;

vector<vector<ll>> dp((1 << 15),vector<ll>(15,-1));
vector<vector<pair<ll, ll>>> graph(15);
ll n;

ll solve(ll bit, ll v)
{
    if(dp.at(bit).at(v) != -1){
        return dp.at(bit).at(v);
    }
    if(bit == (1 << n) -1 && v == 0){
        return dp.at(bit).at(v) = 0;
    }
    ll ans = 1e9+7;
    ll i,j;
    for(i = 0;i < graph.at(v).size();++i){
        ll s = graph.at(v).at(i).first;
        if(bit & (1 << s)) continue;
        ll d = graph.at(v).at(i).second;
        ans = min(ans, solve(bit | 1 << s, s) + d);
    }
    return dp.at(bit).at(v) = ans;
}

int main()
{
    ll i,j;
    ll e;
    cin >> n >> e;
    for(i = 0;i < e;++i){
        ll s,t,d;
        cin >> s >> t >> d;
        graph.at(s).emplace_back(t, d);
    }
    ll ans = min((ll)1e9+7, solve(0LL, 0LL));
    if(ans == 1e9+7){
        cout << -1 << endl;
    }else{
        cout << ans << endl;
    }

    return 0;
}

```