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 ll long long
int n,m,u,v,w;
ll bismarck[(1<<15)][15],e[15][15],sum[(1<<15)];
int main()
{
	cin>>n>>m;
	while(m--)
	{
		cin>>u>>v>>w;
		u--,v--;
		e[u][v]=e[v][u]=w;
	}
	for(int i=0;i<1<<n;i++)
	{
		for(int j=0;j<n;j++){
		if((i>>j)&1)
		{
            for(int k=j+1;k<n;k++)
		    if((i>>k)&1) sum[i]+=e[j][k];			
		}
		bismarck[i][j]=-1e18;}
	}
	bismarck[1][0]=0;
	for(int i=1;i<(1<<n);i++)
	{
		for(int j=0;j<n;j++)
		if((i>>j)&1)
		{
			int l=i^(1<<j);
			for(int k=0;k<n;k++)
			if((i>>k)&1&&e[j][k]) bismarck[i][j]=max(bismarck[i][j],bismarck[l][k]+e[k][j]);
			for(int k=l;k>0;k=(k-1)&l) bismarck[i][j]=max(bismarck[i][j],bismarck[i^k][j]+sum[k^(1<<j)]); 
		}
	}
	cout<<sum[(1<<n)-1]-bismarck[(1<<n)-1][n-1];
}