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>
using namespace std;
using Int = long long;

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}


template<typename T>
vector<T> make_v(size_t a){return vector<T>(a);}

template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts){
  return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}

template<typename T,typename V>
typename enable_if<is_class<T>::value==0>::type
fill_v(T &t,const V &v){t=v;}

template<typename T,typename V>
typename enable_if<is_class<T>::value!=0>::type
fill_v(T &t,const V &v){
  for(auto &e:t) fill_v(e,v);
}

//INSERT ABOVE HERE
signed main(){
  Int n,m;
  cin>>n>>m;
  using P = pair<Int, Int>;
  vector<vector<P> > G(n);
  for(Int i=0;i<m;i++){
    Int s,t,d;
    cin>>s>>t>>d;
    G[s].emplace_back(t,d);
  }
  const Int INF = 1e15;
  Int ans=INF;
  
  for(Int p=0;p<n;p++){
    auto dp=make_v<Int>(n,1<<n);
    fill_v(dp,INF);
    dp[p][1<<p]=0;
    for(Int b=0;b<(1<<n);b++){
      for(Int v=0;v<n;v++){
	if((~b>>v)&1) continue;
	for(auto e:G[v]){
	  Int u,d;
	  tie(u,d)=e;
	  if(u==p&&b==(1<<n)-1) chmin(ans,dp[v][b]+d);
	  if((b>>u)&1) continue;   
	  chmin(dp[u][b|(1<<u)],dp[v][b]+d);
	}
      }
    }
  }
  
  if(ans==INF) ans=-1;
  cout<<ans<<endl;
  return 0;
}

