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;
int f[20][70010],sum[70010];
int mp[20][20];
int main()
{
	int n,m,all=0;scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++)
	{
		int u,v,w;scanf("%d%d%d",&u,&v,&w);
		mp[u][v]=mp[v][u]=w;all+=w;
	}
	memset(f,255,sizeof(f));
	for (int i=1;i<(1<<n);i++)
	for (int j=1;j<=n;j++) if((1<<(j-1))&i)
	for (int k=j+1;k<=n;k++) if((1<<(k-1))&i) sum[i]+=mp[j][k];
	for (int i=1;i<(1<<n);i++) if(i&1) f[1][i]=sum[i];
	for (int j=1;j<(1<<n);j++) 
	{
		for (int i=1;i<=n;i++) if(f[i][j]>=0)
		{
			int k=((1<<n)-1)^j;
			for(int p=k;p;p=(p-1)&k)
				for (int d=1;d<=n;d++)
					if(((1<<(d-1))&p)&&(mp[i][d])) f[d][j|p]=max(f[d][j|p],f[i][j]+mp[i][d]+sum[p]);
		}
	}
	int ans=0;
	for (int i=1;i<(1<<n);i++) ans=max(ans,f[n][i]);
	printf("%d\n",all-ans);
}