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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m;
int f[1<<15][15];
int C[1<<15][15];
int c[15][15];
int cost(int S1,int S2){
  int ans=0;
  for(int i=0;i<n;++i){
    if(S2>>i&1)ans+=C[S1][i];
  }
  return ans;
}
int main(){
  scanf("%d%d",&n,&m);
  memset(c,0xc2,sizeof(c));
  for(int i=1,a,b,w;i<=m;++i){
    scanf("%d%d%d",&a,&b,&w);
    a--;b--;
    c[a][b]=c[b][a]=w;
  }
  int lim=1<<n;
  for(int i=0;i<lim;++i){
    for(int j=0;j<n;++j){
      for(int k=0;k<n;++k){
	if((i>>j&1)&&(!(i>>k&1))&&c[j][k]!=0xc2c2c2c2){
	  C[i][k]+=c[j][k];
	}
      }
    }
  }
  memset(f,0x3f,sizeof(f));
  f[1][0]=0;
  int ans=0x3f3f3f3f;
  for(int i=1;i<lim;++i){
    for(int j=0;j<n;++j){
      if(f[i][j]==0x3f3f3f3f)continue;
      for(int k=0;k<n;++k){
	if((!(i>>k&1))&&c[j][k]!=0xc2c2c2c2){
	  f[i|(1<<k)][k]=min(f[i|(1<<k)][k],f[i][j]+cost(i^(1<<j),1<<k));
	}
      }
      int S=(lim-1)^i;
      for(int j2=S;j2;j2=(j2-1)&S){
	f[i|j2][j]=min(f[i|j2][j],f[i][j]+cost(i^(1<<j),j2));
      }
    }
  }
  printf("%d\n",f[lim-1][n-1]);
  return 0;
}