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;
typedef long long ll;
const int N=1<<15;
int n,m,c[15][15];
int p[N],dp[N][15];
void init()
{
	memset(p,0,sizeof(p));
	for(int i=0;i<(1<<n);i++)
		for(int j=0;j<n;j++)
			dp[i][j]=-1;
	for(int i=0;i<(1<<n);i++)
	{
		for(int j=0;j<n;j++)
		{
			if(((i>>j)&1)==0)
				continue;
			for(int k=j+1;k<n;k++)
			{
				if(((i>>k)&1)==0)
					continue;
				p[i]+=c[j][k];
			}
		}
	}
}
int main()
{
	while(cin>>n>>m)
	{
		int sum=0; 
		int u,v,w;
		memset(c,0,sizeof(c));
		while(m--)
		{
			scanf("%d%d%d",&u,&v,&w);
			u--,v--;
			c[u][v]=c[v][u]=w;
			sum+=w;
		}

        
		init();
		dp[1][0]=0;
		for(int i=0;i<(1<<n);i++)
		{
			for(int j=0;j<n;j++)
			{
				if(dp[i][j]==-1)
					continue;
				int x=(((1<<n)-1)^i)|(1<<j);	//y is some subset of x
				for(int y=x;y>=0;y--)
					y&=x,dp[i|y][j]=max(dp[i|y][j],dp[i][j]+p[y]);	
					
				for(int k=0;k<n;k++)
				{
					if((i>>k)&1)
						continue;
					if(c[j][k]==0)
						continue;
					dp[i|(1<<k)][k]=max(dp[i|(1<<k)][k],dp[i][j]+c[j][k]);
				}
			}
		}
		int ans=1e9;
		for(int i=0;i<(1<<n);i++)
			ans=min(ans,sum-dp[i][n-1]);
		cout<<ans<<endl;
	}
	return 0;
}