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;
int n,m;
int add[1<<15];
int dp[1<<15][15];
int i,j,k,p;
int x,y,z;
int mp[15][15];
int main()
{
	cin>>n>>m;
	while (m--)
	{
		cin>>x>>y>>z;
		mp[x-1][y-1]=z;
		mp[y-1][x-1]=z;
	}
	for (i=0;i<(1<<n);i++)
	{
		for (j=0;j<n;j++)
		{
			if ((1<<j)&i)
			{
				for (k=j+1;k<n;k++)
				{
					if ((1<<k)&i)
					{
						//cout<<j<<' '<<k<<endl;
						add[i]+=mp[k][j];
					}
				}
			}
		}
	}
	memset(dp,-1,sizeof(dp));
	dp[1][0]=0;
	for (i=1;i<(1<<n);i++)
	{
		for (j=0;j<n;j++)
		{
			if ((1<<j)&i)
			{
				if (dp[i][j]!=-1)
				{
					for (k=0;k<n;k++)
					{
						if (!((1<<k)&i))
						{
							if (mp[j][k])
							{
								dp[i|(1<<k)][k]=max(dp[i|(1<<k)][k],dp[i][j]+mp[j][k]);
							}
						}
					}
					int bj=((((1<<n)-1)^i)|(1<<j));
					for (k=bj;;k=((k-1)&bj))
					{
						dp[i|k][j]=max(dp[i|k][j],dp[i][j]+add[k]);
						if (!k) break;
					}
				}
			}
		}
	}
	int ans=0;
	for (i=1;i<(1<<n);i++) ans=max(ans,dp[i][n-1]);
	//cout<<add[(1<<n)-1]<<endl; 
	cout<<add[(1<<n)-1]-ans<<endl;
	return 0;
}