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<cstdio>
using namespace std;
int dp[66666][17][17],n,m,mp[17][17],xx,yy,zz,k,sum;
int main()
{
	cin>>n>>m;
	for (int i=1;i<=m;i++)
	{
		cin>>xx>>yy>>zz;
		xx--;yy--;
		mp[xx][yy]=zz;
		mp[yy][xx]=zz;
	}
	for (int i=0;i<(1<<n);i++)
	{
		for (int j=0;j<n;j++)
		{
			for (int k=0;k<n;k++)
			dp[i][j][k]=1e9;
		}
		for (int j=0;j<n;j++) dp[i][j][j]=0;
	}
	for (int i=1;i<(1<<n);i++)
	{
		for (int j=i;j;j=(j-1)&i)
		{
			if (!(j&1)) continue;
			k=i^j;
			sum=0;
			for (int x=0;x<n;x++)
			{
				if (j&(1<<x)) 
				{
					for (int y=0;y<n;y++)
					{
						if (k&(1<<y)) sum+=mp[x][y]; 
					}
				}
			}
			int x=0;
			for (int y=0;y<n;y++)
			{
				if (k&(1<<y))
				{
					for (int z=0;z<n;z++)
					{
						if (j&(1<<z) && mp[y][z]) 
						{
							dp[i][x][y]=min(dp[i][x][y],dp[j][x][z]+sum-mp[y][z]);
						}
					}
				}
			}
		}
	}
	cout<<dp[(1<<n)-1][0][n-1]<<endl;
	return 0;
}