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 N 17
using namespace std;
int n,m;
int mp[N][N],sum[1<<15],f[1<<15][N];
int main()
{
	scanf("%d%d",&n,&m);int t1,t2,t3;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%d",&t1,&t2,&t3);
		mp[t1][t2]=mp[t2][t1]=t3;
	}
	for(int i=0;i<1<<n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			if(i&(1<<(j-1)))
			{
				for(int k=j+1;k<=n;k++)
				{
					if(i&(1<<(k-1)))
					{
						sum[i]+=mp[j][k];
					}
				}
			}
		}
	}
	memset(f,0xcf,sizeof(f));
	for(int i=1;i<1<<n;i++)if(i&1)f[i][1]=sum[i];
	for(int i=1;i<1<<n;i++)
	{
		for(int k=1;k<=n;k++)
		{
			if(i&(1<<(k-1)))
			{
				for(int j=(i-1)&i;j;j=(j-1)&i)
				{
					if(j&(1<<(k-1)))continue;
					for(int l=1;l<=n;l++)
					{
						if((j&(1<<(l-1)))&&mp[l][k])
						{
							f[i][k]=max(f[i][k],f[j][l]+mp[l][k]+sum[i^j]);
						}
					}
				}
			}
		}
	}
	printf("%d\n",sum[(1<<n)-1]-f[(1<<n)-1][n]);
	return 0;
}