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<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
using namespace std;
int n,m,dis[20][20],id[1<<16],sum=0,p[1<<16],f[1<<16][20],ans;
const int inf=1<<28;
int main()
{
	scanf("%d %d",&n,&m);
	for(int i=1;i<=n;i++) id[1<<(i-1)]=i;
	for(int i=1;i<=m;i++)
	{
		int x,y,c;scanf("%d %d %d",&x,&y,&c);
		dis[x][y]=c;dis[y][x]=c;sum+=c;
	}
	for(int i=1;i<1<<n;i++)
	{
        int x=id[i&(-i)];
        p[i]=p[i^(1<<(x-1))];
        for(int j=x+1;j<=n;j++) if(i&(1<<(j-1))) p[i]+=dis[x][j];
    }
    for(int i=0;i<=n;i++) for(int j=0;j<1<<n;j++) f[j][i]=-inf;
    f[1][1]=0;ans=-inf;
    for(int s=1;s<1<<n;s++)
	{
        for(int i=1;i<=n;i++)
		{
            int r=s;
            while(r)
			{
                if((r&(1<<(i-1)))) f[s][i]=max(f[s][i],f[(s-r)|(1<<(i-1))][i]+p[r]);
                r=(r-1)&s;
            }
            for(int j=1;j<=n;j++)
                if (!(s&(1<<(j-1)))&&dis[i][j]) f[s|(1<<(j-1))][j]=max(f[s|(1<<(j-1))][j],f[s][i]+dis[i][j]);
        }
        if((s&(1<<(n-1)))) ans=max(ans,f[s][n]);
    }
    printf("%d",sum-ans);
}