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<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<complex>
#include<iostream>
#include<algorithm>
#define ll long long
#define inf 1e9
using namespace std;

inline void up(int &x,const int &y){if(x<y)x=y;}
const int maxn = 20;
const int maxm = 81000;

int n,m;
int a[maxn][maxn];
int ct[maxm],f[maxm][maxn];

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int x,y,c; scanf("%d%d%d",&x,&y,&c);
		a[x][y]=a[y][x]=c;
	}
	int al=1<<n;
	for(int i=1;i<al;i++)
	{
		int j=1;for(;j<=n;j++) if(i>>j-1&1)break;
		ct[i]=ct[i^1<<j-1];
		for(int k=1;k<=n;k++) if(i>>k-1&1) ct[i]+=a[k][j];
	}
	for(int i=0;i<al;i++) for(int j=1;j<=n;j++) f[i][j]=-inf;
	
	f[1][1]=0; int re=0;
	for(int i=1;i<al;i+=2)
	{
		for(int j=1;j<=n;j++) if((i>>j-1&1)&&f[i][j]>=0)
		{
			if(j==n) up(re,f[i][j]);
			for(int k=1;k<=n;k++) if(!(i>>k-1&1)&&a[j][k])
				up(f[i|1<<k-1][k],f[i][j]+a[j][k]);
			
			int now=al-1-i;
			for(int k=now;k;k=k-1&now)
				up(f[i|k][j],f[i][j]+ct[k^1<<j-1]);
		}
	}
	printf("%d\n",ct[al-1]-re);
	
	return 0;
}