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;
#define MAXN (1<<15)+3
int n,m,a,b,c,sum[MAXN],f[MAXN][21],p[21][21],res,ans;
int main(){
	scanf("%d%d",&n,&m);
	int maxs=(1<<n)-1;
	for(int i=1;i<=m;i++){
		scanf("%d%d%d",&a,&b,&c);
		p[a][b]=p[b][a]=c;
		res+=c;
	}
	for(int s=0;s<=maxs;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))){
			  	sum[s]+=p[i][j];
			  }
		}
		sum[s]>>=1;
	}
	memset(f,-1,sizeof(f));
	f[1][1]=0;
	for(int s=0;s<=maxs;s++){
		for(int i=1;i<=n;i++){
			if(f[s][i]==-1) continue;
			for(int j=1;j<=n;j++){
				if(s&(1<<(j-1))) continue;
				if(!p[i][j]) continue;
				f[s|(1<<(j-1))][j]=max(f[s|(1<<(j-1))][j],f[s][i]+p[i][j]);
			}
			int rest=(maxs^s)|(1<<(i-1));
			for(int ss=rest;ss;ss=rest&(ss-1)){
				f[s|ss][i]=max(f[s|ss][i],f[s][i]+sum[ss]);
			}
		}
	}
	int ans=0;
	for(int i=0;i<=maxs;i++)
	  ans=max(ans,f[i][n]);
	printf("%d\n",res-ans);
}