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 n,m,i,j,x,y,w,hd,s,xlh,g[100010],v[21][21],f[100010][21],sy,mx,ans,pr,sz;
int main(){
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;i++){
		scanf("%d%d%d",&x,&y,&w);
		xlh+=w;
		x--;y--;
		v[x][y]=v[y][x]=w;
	}
	mx=1<<n;
	for(s=1;s<mx;s++){
		for(i=0;i<n;i++)if(s&(1<<i)){
			pr=s^(1<<i);
			sz=0;
			for(j=0;j<n;j++)if((s&(1<<j))&&v[i][j])sz+=v[i][j];
			g[s]=g[pr]+sz;break;
		}
		//printf("%d %d\n",s,g[s]);
	}
	memset(f,172,sizeof(f));
	f[1][0]=0;
	for(s=1;s<mx;s++)
	 for(hd=0;hd<n;hd++)if(f[s][hd]>=0){
	 	for(i=0;i<n;i++)if(!(s&(1<<i))&&v[hd][i]){
	 		f[s|(1<<i)][i]=max(f[s|(1<<i)][i],f[s][hd]+v[hd][i]);
		 }
		sy=(mx-1)^s;
		for(i=sy;i;i=(i-1)&sy)f[s|i][hd]=max(f[s|i][hd],f[s][hd]+g[i|(1<<hd)]);
	 }
	 for(i=1;i<mx;i++)ans=max(ans,f[i][n-1]);
	 printf("%d",xlh-ans);
}