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<cstring>
#include<algorithm>
#define MN 16
using namespace std;
inline int in(){
    int x=0;bool f=0; char c;
    for (;(c=getchar())<'0'||c>'9';f=c=='-');
    for (x=c-'0';(c=getchar())>='0'&&c<='9';x=(x<<3)+(x<<1)+c-'0');
    return f?-x:x;
}
int f[1<<MN][MN],sum[1<<MN],mp[MN][MN];
int n,m,a,b,c;
int main()
{
	n=in();m=in();
	for (int i=1;i<=m;++i){
		a=in()-1;b=in()-1;c=in();
		mp[a][b]=mp[b][a]=c;
	}memset(f,-0X3F,sizeof(f));f[1][0]=0;
	for (int s=0;s<(1<<n);++s)
	for (int i=0;i<n;++i) if (s&(1<<i))
	for (int j=i+1;j<n;++j) if (s&(1<<j)) sum[s]+=mp[i][j];
	for (int s=0;s<(1<<n);++s)
	for (int i=0;i<n;++i) if (s&(1<<i)){
		for (int j=0;j<n;++j)
		if (!(s&(1<<j))&&mp[i][j])
		f[s|(1<<j)][j]=max(f[s|(1<<j)][j],f[s][i]+mp[i][j]);
		int r=((1<<n)-1)-s;
		for (int t=r;t;t=((t-1)&r)) f[s|t][i]=max(f[s|t][i],f[s][i]+sum[t+(1<<i)]);
	}
	printf("%d",sum[(1<<n)-1]-f[(1<<n)-1][n-1]);return 0;
}