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;
inline int gi(){
  char ch=getchar();int x=0,q=0;
  while(ch<'0' || ch>'9') ch=='-'?q=1:0,ch=getchar();
  while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
  return q?(-x):x;
}
int f[35000][17],w[35000],c[17][17];
void dp(int &x,int y){
  if (x<y) x=y;
  return;
}
int main(){
  int n=gi(),m=gi(),z,a,b,p,S=(1<<n)-1,d=0;
  for (int i=1; i<=m; ++i){
    a=gi(),b=gi(),p=gi();
    c[a][b]+=p,c[b][a]+=p;
    z=(1<<(a-1))|(1<<(b-1));
    for (int s=1; s<(1<<n); ++s)
      if ((s&z)==z)
	w[s]+=p;
    d+=p;
  }
  memset(f,-100,sizeof(f));
  f[1][1]=0;
  for (int s=1; s<=S; ++s)
    for (int i=1; i<=n; ++i)
      if (s&(1<<(i-1))){
	for (int j=1; j<=n; ++j)
	  if (!(s&(1<<(j-1)))&&c[i][j])
	    dp(f[s|(1<<(j-1))][j],f[s][i]+c[i][j]);
	for (int t=S^s; t; t=(t-1)&(S^s))
	  dp(f[s|t][i],f[s][i]+w[t|1<<(i-1)]);
      }
  int ans=0;
  for (int s=1; s<=S; ++s)
    dp(ans,f[s][n]);
  cout<<d-ans;
  return 0;
}