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<iostream>
#include<cstdio>
using namespace std;
int n,m;
int dp[1<<16][16];
int sum[1<<16];
int dis[16][16];
int all;
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		x--;y--;
		dis[x][y]=dis[y][x]=z;
		all+=z;
	}
	for(int i=1;i<(1<<n);i++)
	{
		for(int j=0;j<n;j++)
		{
			if(i>>j&1)
			{
				for(int k=j+1;k<n;k++)
				{
					if(i>>k&1)
					{
						sum[i]+=dis[j][k];
					}
				}
			}
		}
	}
	for(int i=0;i<(1<<n);i++)
	{
		for(int j=0;j<n;j++)dp[i][j]=-1;
	}
	dp[1][0]=0;
	for(int i=1;i+1<(1<<n);i++)
	{
		for(int j=0;j<n;j++)
		{
			if(dp[i][j]==-1)continue;
			int st=((1<<n)-1)^i;
			if(dis[j][n-1])
			{
				dp[(1<<n)-1][n-1]=max(dp[(1<<n)-1][n-1],dp[i][j]+sum[st]+dis[j][n-1]);
			}
			for(int k=(st-1)&st;k;k=(k-1)&st)
			{
				if(k>>(n-1)&1)continue;
				dp[i|k][j]=max(dp[i|k][j],dp[i][j]+sum[(1<<j)|k]);
			}
			for(int k=0;k<n;k++)
			{
				if(k!=n-1 && !(i>>k&1) && dis[j][k])
				{
					dp[i|(1<<k)][k]=max(dp[i|(1<<k)][k],dp[i][j]+dis[j][k]);
				}
			}
		}
	}
	printf("%d",all-dp[(1<<n)-1][n-1]);
	return 0;
}