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;
int read()
{
	int x=0,f=1;	char ch=getchar();
	while(ch<'0'||ch>'9')	{if(ch=='-')	f=-1;	ch=getchar();}
	while(ch>='0'&&ch<='9')	{x=(x<<3)+(x<<1)+(ch^48);	ch=getchar();}
	return x*f;
}
int n,m,x,y,z;
int dis[20][20];
int tag[100050],s[100050];
int f[100050][20],sum,ans;
int main()
{
	n=read();	m=read();	tag[0]=-1;
	memset(f,-1,sizeof(f));	f[1][0]=0;
	while(m--)
	{
		x=read()-1;	y=read()-1;	z=read();
		dis[x][y]=dis[y][x]=z;	sum+=z;
	}
	for(int i=1;i<(1<<n);++i)
	{
		x=tag[i]=tag[i>>1]+1;	s[i]=s[i^(1<<x)];
		for(y=0;y<x;++y)
			if(i&(1<<y))	s[i]+=dis[x][y];
	}
	for(int i=1;i<(1<<n);++i)
		for(int j=0;j<n;++j)
		{
			if(f[i][j]==-1)	continue;
			for(int k=0;k<n;++k)
				if(dis[j][k]&&!(i&(1<<k)))
					f[i|(1<<k)][k]=max(f[i|(1<<k)][k],f[i][j]+dis[j][k]);
			int t=((1<<n)-1)^i^(1<<j);
			for(int k=t;k;k=(k-1)&t)
				f[i|k][j]=max(f[i|k][j],f[i][j]+s[k]);
		}
	for(int i=0;i<(1<<n);++i)
		ans=max(ans,f[i][n-1]);
	printf("%d",sum-ans);
	return 0;
}