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 maxn=15;
const int maxa=1<<15;
const int inf=0x3f3f3f3f;
int n,m,f[maxn][maxa],g[maxa],e[maxn][maxn];
int main()
{
	scanf("%d%d",&n,&m);
	int x,y,w,sum=0;
	for(int i=0;i<m;i++)
		scanf("%d%d%d",&x,&y,&w),x--,y--,e[x][y]=e[y][x]=w,sum+=w;
	memset(g,inf,sizeof(g));
	g[0]=0;
	for(int i=0;i<n;i++)
		for(int j=0;j<(1<<n);j++)
			if(g[j]!=inf && !(j&(1<<i)))
			{
				int val=0;
				for(int k=0;k<n;k++)
					if(j&(1<<k)) val+=e[i][k];
				g[j|(1<<i)]=g[j]+val;
			}
	memset(f,-1,sizeof(f));
	for(int i=0;i<(1<<n);i++)
		if(i&1) f[0][i]=g[i];
	for(int i=1;i<(1<<n);i++)
		for(int j=0;j<n;j++)
			if(i&(1<<j) && f[j][i]!=-1)
				for(int k=i^((1<<n)-1),t=k;k>=0;k=k?(k-1)&t:-1)
					for(int l=0;l<n;l++)
						if(!(i&(1<<l)) && (k&(1<<l)) && e[j][l])
							f[l][i|k]=max(f[l][i|k],f[j][i]+g[k]+e[j][l]);
	printf("%d\n",sum-f[n-1][(1<<n)-1]);
	return 0;
}