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;
#define N 15

int n,m,sum,d[N][1<<N],e[1<<N],f[1<<N][N];

void upd(int &x,int y){x=max(x,y);}

int read(){
	int x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-')  f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

int main(){
	n=read();m=read();
	for (int i=1;i<=m;i++){
		int x=read()-1,y=read()-1;
		d[x][1<<y]=d[y][1<<x]=read();
		sum+=d[x][1<<y];
	}
	for (int i=0;i<n;i++)
		for (int j=1;j<(1<<n);j++)
			d[i][j]=d[i][j&-j]+d[i][j-(j&-j)];
	for (int i=0;i<(1<<n);i++)
		for (int j=0;j<n;j++)
			if (i&(1<<j))
			e[i]=e[i-(1<<j)]+d[j][i-(1<<j)];
	memset(f,0x80,sizeof(f));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 ((i&(1<<k))==0&&d[j][1<<k])
					upd(f[i+(1<<k)][k],f[i][j]+d[j][1<<k]);
				for (int s=(1<<n)-1-i,k=s;k;k=(k-1)&s)
					upd(f[i+k][j],f[i][j]+e[k+(1<<j)]);
			}
	printf("%d\n",sum-f[(1<<n)-1][n-1]);
	return 0;
}