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
#include <bits/stdc++.h>
 
#define rep(i,n) for(ll i=0;i<n;++i)
#define ALL(a) (a).begin(),(a).end()
 
using namespace std;
using ll  = long long;
using pll = pair<ll, ll>;
 
const ll MOD  = 1e9+7;
const ll LINF = 1LL << 60;
const int INF = 1e9 + 7;

int main(){
    int v, e;
    cin >> v >> e;

    ll dist[v][v];//dist[i][j]:iとjを結ぶ辺の距離、
    rep(i, v)rep(j, v)dist[i][j] = INF;

    rep(i, e){
        int s, t, d;
        cin >> s >> t >> d;

        dist[s][t] = d;
    }

    ll dp[(1<<v)][v];//dp[r][i]:集合r(立っているbitの番号の辺が訪問済)から頂点iへ行くときの最短距離
    rep(i, (1<<v))rep(j, v)dp[i][j] = INF;

    dp[0][0] = 0;
    rep(r, (1<<v)){
        rep(i, v){
            if(r&(1<<i)){
                rep(j, v)dp[r][i] = min(dp[r][i], dp[r-(1<<i)][j]+dist[j][i]);
            }
        }
    }

    ll ans = dp[(1<<v)-1][0];
    if(ans >= INF)cout << -1 << endl;
    else cout << ans << endl;
}
