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>
#define ll long long
using namespace std;
int ha,n,m,i,j,x,y,w,k,mx,ans,sz[100010],f[50001][16],a[101][101];
int main(){
	scanf("%d%d",&n,&m);
	memset(f,-1,sizeof(f));
	f[1][0]=0;
	for(i=1;i<=m;i++){
		scanf("%d%d%d",&x,&y,&w);
		x--;y--;
		a[x][y]=a[y][x]=w;
		ans+=w;
	}
	mx=1<<n;
	for(i=0;i<mx;i++){
		for(j=0;j<n;j++)if(i&(1<<j))
		 for(k=j+1;k<n;k++)if(i&(1<<k))sz[i]+=a[j][k];
	}
	for(i=1;i<mx-1;i++)
	 for(j=0;j<n;j++)if(f[i][j]!=-1){
	 //	printf("%d %d %d\n",i,j,f[i][j]);
	 	for(k=0;k<n;k++)if((!(i&(1<<k)))&&a[j][k])f[i|1<<k][k]=max(f[i|1<<k][k],f[i][j]+a[j][k]);
	 	for(k=(mx-1)^i;k;k=(k-1)&((mx-1)^i))if(!(k&(1<<n-1)))f[i|k][j]=max(f[i|k][j],f[i][j]+sz[k|1<<j]);
	 }
	//printf("%d\n",mx);
	for(i=1;i<mx;i++){
	  ha=max(ha,f[i][n-1]);
	  //printf("%d %d %d\n",i,n-1,f[i][n-1]);
}
	printf("%d",ans-ha);
}
/*
一道挺好想的dp，
最后的路径一定是一条从s到t的链和在链上的点向外挂出去的一些联通分量。
那么就可以dp表示到了i与s联通的集合为j 
*/