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<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<iostream>
#define N 20
#define M (1<<15)+10
#define lowbit(x) (x&(-x))
using namespace std;
int f[M][N],g[M][N],h[M],a[N][N],bit[M],n,m,tot,ans;
void upd(int &x,int y)
{
	if(y>x) x=y;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int x,y,d;scanf("%d%d%d",&x,&y,&d);
		tot+=d;
		a[x][y]=a[y][x]=d;
	}
	for(int i=1;i<=n;i++) bit[1<<(i-1)]=i;
	memset(f,-1,sizeof(f));
	memset(g,-1,sizeof(g));
	int ss=1<<n;
	for(int i=1;i<ss;i++)
	{
		int j=bit[lowbit(i)];
		h[i]=h[i-lowbit(i)];
		for(int k=1;k<=n;k++) if(j!=k && (i&(1<<(k-1)))) h[i]+=a[j][k];
	}
	f[1][1]=0;
	for(int i=0;i<ss;i++)
		for(int j=1;j<=n;j++)
		{
			upd(g[i][j],f[i][j]);
			if(f[i][j]!=-1)
			{
				int t=ss-1-i;
				for(int k=t;k;k=(k-1)&t)
					upd(g[i+k][j],f[i][j]+h[k+(1<<(j-1))]);	
			}
			if(g[i][j]!=-1)
			{
				for(int k=1;k<=n;k++)
				{
					if((i&(1<<(k-1))) || a[j][k]==0) continue;
					upd(f[i+(1<<(k-1))][k],g[i][j]+a[j][k]);
				}
			}
		}
	ans=max(f[ss-1][n],g[ss-1][n]);
	printf("%d\n",tot-ans);
	return 0;
}