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 maxn 100005
#define ll long long
using namespace std;
int n,m;
int g[16][16];
int block[1<<15],f[1<<15][16];
int main()
{
	scanf("%d%d",&n,&m);
	int u,v,w,sum=0;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%d",&u,&v,&w);
		sum+=w;g[u][v]=g[v][u]=w;
	}
	for(int s=0;s<(1<<n);s++)
	{
		for(int j=1;j<=n;j++)
		{
			if((!(s>>(j-1))&1)&&block[s|(1<<(j-1))]==0)
			{
				block[s|(1<<(j-1))]=block[s];
				for(int k=1;k<=n;k++)
				{
					if((s>>(k-1))&1)
						block[s|(1<<(j-1))]+=g[j][k];
				}
			}
		}
	}
	memset(f,-1,sizeof(f));
	f[1][1]=0;
	for(int s=0;s<(1<<n);s++)
	{
		for(int j=1;j<=n;j++)
		{
			if(f[s][j]!=-1&&((s>>(j-1))&1))
			{
				for(int k=1;k<=n;k++)
				{
					if(!((s>>(k-1))&1)&&g[j][k])
						f[s|(1<<(k-1))][k]=max(f[s|(1<<(k-1))][k],f[s][j]+g[j][k]);
				}
				int ss=(((1<<n)-1)^s)|(1<<(j-1));
				for(int k=ss;k;k=(k-1)&ss)
				{
					if((k>>(j-1))&1)
						f[s|k][j]=max(f[s|k][j],f[s][j]+block[k]);
				}
			}
		}
	}
	printf("%d\n",sum-f[(1<<n)-1][n]);
	return 0;
}