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:
/*********************
*  Author: xuziyuan  * 
*********************/
 
#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define LL long long
#define pii pair <int,int>
#define fi first
#define se second
#define pb push_back
#define mpr make_pair
 
using namespace std;
 
const LL MOD=1e9+7;

int n,m,x,y,z,a[20][20],val[40000],dp[40000][20],sum=0;

int main()
{
	cin>>n>>m;
	rep(i,m)
	{
		cin>>x>>y>>z;
		x--;
		y--;
		sum+=z;
		a[x][y]=a[y][x]=z;
	}
	rep(i,1<<n) rep(j,n) for(int k=j+1;k<n;k++) if((i&(1<<j))>0&&(i&(1<<k))>0) val[i]+=a[j][k];
	rep(i,39995) rep(j,18) dp[i][j]=-1;
	rep(i,1<<n) if((i&1)==1) dp[i][0]=val[i];
	rep(i,1<<n)
	{
		rep(j,n)
		{
			if(dp[i][j]==-1) continue;
			int rest=(1<<n)-1-i;
			for(int k=rest;k>0;k=(k-1)&rest)
			{
				rep(p,n)
				{
					if((k&(1<<p))==0||a[j][p]==0) continue;
					dp[i|k][p]=max(dp[i|k][p],dp[i][j]+val[k]+a[j][p]);
				}
			}
		}
	}
	cout<<sum-dp[(1<<n)-1][n-1]<<endl;
	return 0;
}