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 cmax(a,b) a=max(a,b)
using namespace std;
typedef long long ll;
const ll N=20,V=(1ll<<15)+20,B=(ll)1e15;
ll n,m,sx,S=0,ans;
ll f[N][V],h[V],e[N][N];
int main()
{
	ll i,j,ns,vs,hsx,u,v,w;
	scanf("%lld%lld",&n,&m);
	sx=1<<n;
	for(i=1;i<=m;i++)
	{
		scanf("%lld%lld%lld",&u,&v,&w);u--;v--;
		e[u][v]=e[v][u]=w;
		for(ns=0;ns<sx;ns++)
		{
			if((ns&(1<<u))&&(ns&(1<<v))) h[ns]+=w;	
		}
		S+=w;
	}
	memset(f,-0x3f,sizeof(f));
	f[0][1]=0;
	for(ns=1;ns<sx;ns++)
	{
		for(i=0;i<n;i++)
		{
			if((ns&(1<<i))==0) continue;
			if(f[i][ns]<-B) continue;
			//printf("#%lld %lld  %lld\n",i,ns,f[i][ns]);
			hsx=(sx-1)^ns;
			if(hsx>0)
			{
				for(vs=hsx;vs;vs=(vs-1)&hsx)
				{
					cmax(f[i][ns|vs],f[i][ns]+h[vs|(1<<i)]);
				}
			}
			for(j=0;j<n;j++)
			{
				if((ns&(1<<j))||e[i][j]==0) continue;
				cmax(f[j][ns|(1<<j)],f[i][ns]+e[i][j]);
			}
		}
	}
	//printf("P%lld\n",f[n-1][sx-1]);
	ans=S-f[n-1][sx-1];
	printf("%lld",ans);
	return 0;
}