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<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
using namespace std;

typedef long long LL;

int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}

inline void upd(int &x,int y) { x<y?x=y:0; }

int f[1<<15][15];
int sum[1<<15];
int e[15][15];
int main()
{
	int n=gi(),m=gi(),i,a,b,s,t,o,k;
	const int all=(1<<n)-1;
	
	while (m--) {
		a=gi()-1,b=gi()-1;
		e[a][b]=e[b][a]=gi();
		sum[1<<a|1<<b]+=e[a][b];
	}

	for (i=0;i<n;i++)
		for (s=0;s<(1<<n);s++)
			if (!(s>>i&1))
				sum[s|1<<i]+=sum[s];

	memset(f,-0x3f,sizeof(f));
	f[1][0]=0;
	for (s=1;s<(1<<n);s+=2)
		for (i=0;i<n;i++)
			if (f[s][i]>=0) {
				for (o=all^s,t=o;t;(--t)&=o)
					upd(f[s|t][i],f[s][i]+sum[t|1<<i]);
				
				for (k=0;k<n;k++)
					if (!(s>>k&1)&&e[i][k])
						upd(f[s|1<<k][k],f[s][i]+e[i][k]);
			}
	
	printf("%d\n",sum[all]-f[all][n-1]);
	return 0;
}