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 <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>
#include <ctime>
#include <map>
#include <queue>
#include <cstdlib>
#include <string>
#include <climits>
#include <set>
#include <vector>
using namespace std;
inline int read(){
	int k=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){k=k*10+ch-'0';ch=getchar();}
	return k*f;
}
int n,m,a[110][110],f[1<<15][20],s[1<<15];
int main()
{
	n=read();m=read();int sum=0;
	for(int i=1;i<=m;i++){
		int x=read()-1,y=read()-1,z=read();
		a[x][y]=a[y][x]=z;sum+=z;
	}
	for(int i=0;i<(1<<n);i++)
		for(int j=0;j<n;j++)f[i][j]=-1;
	for(int i=1;i<(1<<n);i++)
		for(int j=0;j<n;j++)if((1<<j)&i)
			for(int k=j+1;k<n;k++)if((1<<k)&i)s[i]+=a[j][k];
	f[1][0]=0;
	for(int i=1;i<(1<<n);i++)
		for(int j=0;j<n;j++)if(f[i][j]!=-1){
			for(int k=0;k<n;k++)if(((1<<k)&i)==0&&a[j][k]>0)f[i|(1<<k)][k]=max(f[i|(1<<k)][k],f[i][j]+a[j][k]);
			int x=(((1<<n)-1)^i)|(1<<j);
			for(int y=x;y>=0;y--)y&=x,f[i|y][j]=max(f[i|y][j],f[i][j]+s[y]);
		}
	int ans=1e9;
	for(int i=0;i<(1<<n);i++)ans=min(ans,sum-f[i][n-1]);
	printf("%d",ans);
	return 0;
}