Problem: Mole decided to live in an abandoned mine. The structure of the mine is represented by a simple connected undirected graph which consists of N vertices numbered 1 through N and M edges. The i-th edge connects Vertices a_i and b_i, and it costs c_i yen (the currency of Japan) to remove it.

Mole would like to remove some of the edges so that there is exactly one path from Vertex 1 to Vertex N that does not visit the same vertex more than once. Find the minimum budget needed to achieve this.

Constraints

* 2 \leq N \leq 15
* N-1 \leq M \leq N(N-1)/2
* 1 \leq a_i, b_i \leq N
* 1 \leq c_i \leq 10^{6}
* There are neither multiple edges nor self-loops in the given graph.
* The given graph is connected.

Input

Input is given from Standard Input in the following format:


N M
a_1 b_1 c_1
:
a_M b_M c_M


Output

Print the answer.

Examples

Input

4 6
1 2 100
3 1 100
2 4 100
4 3 100
1 4 100
3 2 100


Output

200


Input

2 1
1 2 1


Output

0


Input

15 22
8 13 33418
14 15 55849
7 10 15207
4 6 64328
6 9 86902
15 7 46978
8 14 53526
1 2 8720
14 12 37748
8 3 61543
6 5 32425
4 11 20932
3 12 55123
8 2 45333
9 12 77796
3 9 71922
12 15 70793
2 4 25485
11 6 1436
2 7 81563
7 11 97843
3 1 40491


Output

133677

Solution (in CPP):

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

int N,M;
int G[15][15];

int dp[15][(1<<15)];
int INF=2e9;

int calc(int bit,int nbit){
  int sum=0;
  for(int i=0;i<N;i++)
    if(nbit>>i&1)
      for(int j=0;j<N;j++)
        if(~bit>>j&1)
          if(~nbit>>j&1)
            sum+=G[i][j];
  return sum;
}



int rec(int pos,int bit){
  static unordered_map< int , int > mem[15];
  if( mem[pos].count(bit) )return mem[pos][bit];
  
  if(pos==N-1)return 0;
  int res=INF;

  int sup=(1<<(N-1))-1-bit;
  
  int nbit=sup;
  while(1){

    if(nbit>>pos&1){
      int cost=calc(bit,nbit);
      for(int to=0;to<N;to++){
        if(bit>>to&1)continue;
        if(nbit>>to&1)continue;
        if(G[pos][to]==0)continue;
        res=min(res, rec(to,bit|nbit) + cost - G[pos][to]);
      }
    }
    nbit--;
    nbit&=sup;
    if(nbit==0)break;
  }
  return mem[pos][bit]=res;
}

int main(){
  cin>>N>>M;
  for(int i=0;i<M;i++){
    int a,b,c;
    cin>>a>>b>>c;
    a--,b--;
    G[a][b]=G[b][a]=c;
  }
  cout<< rec(0,0) <<endl;
  return 0;
}