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;
const int N=15;
int d[N][N],dp[1<<N][N],a[1<<N],n,m,ans,sum;
void chmax(int &a,int b){if(a==-1||a<b)a=b;}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++)
	{
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		u--;
		v--;
		d[u][v]=d[v][u]=w;
		sum+=w;
	}
	memset(dp,-1,sizeof(dp));
	for(int s=0;s<(1<<n);s++)
		for(int i=0;i<n;i++)
			if(s&(1<<i))
				for(int j=i+1;j<n;j++)
					if(s&(1<<j))
						a[s]+=d[i][j];
	dp[1][0]=0;
	for(int s=1;s<(1<<n);s++)
	{
		for(int i=0;i<n;i++)
		{
			if(dp[s][i]==-1)
				continue;
			for(int j=0;j<n;j++)
			{
				if(s&(1<<j))
					continue;
				if(!d[i][j])
					continue;
				chmax(dp[s|(1<<j)][j],dp[s][i]+d[i][j]);
			}
			int k=(((1<<n)-1)^s)|(1<<i);
			for(int t=k;t;t=(t-1)&k)
				chmax(dp[s|t][i],dp[s][i]+a[t]);
		}
	}
	for(int s=0;s<(1<<n);s++)
		ans=max(ans,dp[s][n-1]);
	printf("%d\n",sum-ans);
	return 0;
}