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;

inline int read() {
	int res=0,fh=1;
	char ch=getchar();
	while((ch>'9'||ch<'0')&&ch!='-')ch=getchar();
	if(ch=='-')fh=-1,ch=getchar();
	while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar();
	return fh*res;
}

const int maxn=16;
int n,m,mxs,sum[(1<<16)+12],a[maxn][maxn];
int f[(1<<maxn)][maxn];
inline int Max(int a,int b){return a>b?a:b;}

int main(){
	n=read(),m=read();
	mxs=(1<<n)-1;
	for(int i=1;i<=m;i++){
		int x=read()-1,y=read()-1,z=read();
		a[x][y]=a[y][x]=z;
	}
	for(int i=0;i<=mxs;i++){
		sum[i]=0;
		for(int j=0;j<n;j++)
			if((i&(1<<j)))
				for(int k=j+1;k<n;k++)
					if((i&(1<<k))) sum[i]+=a[j][k];
	}
	memset(f,-1,sizeof f);
	f[1][0]=0;
	for(int i=1;i<=mxs;i+=2){
		for(int j=0;j<n;j++){
			if(f[i][j]==-1) continue;
			if(!(i&(1<<j))) continue;
			for(int k=0;k<n;k++)
				if(a[j][k]&&(!(i&(1<<k))))
					f[i|(1<<k)][k]=Max(f[i|(1<<k)][k],f[i][j]+a[j][k]);
			int c=(mxs-i)|(1<<j);
			for(int k=c;k>=0;k=(k==0?-1:(k-1)&c))
				f[i|k][j]=Max(f[i][j]+sum[k],f[i|k][j]);
		}
	}
	int ans=0;
	for(int i=(mxs>>1)+2;i<=mxs;i+=2)
		ans=Max(ans,f[i][n-1]);
	printf("%d\n",sum[mxs]-ans);
	return 0;
}