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,ma[18][18],m,tot[1<<16],dp[1<<16][16],tt;
void getmax(int &x,int y)
{
	if(y>x)x=y;
}
int main()
{
	memset(ma,-0x3f3f3f,sizeof(ma));
	scanf("%d%d",&n,&m);tt=0;
	for(int i=1;i<=m;++i)
	{
		int x,y,z;scanf("%d%d%d",&x,&y,&z);
		ma[x][y]=ma[y][x]=z;tt+=z;
	}
	for(int i=1;i<(1<<n);++i)
	{
		for(int j=1;j<=n;++j)
		{
			if(i&(1<<(j-1)))
			{
				tot[i]=tot[i^(1<<(j-1))];
				for(int k=j+1;k<=n;++k)
				{
					if((i&(1<<(k-1)))&&ma[j][k]>0)tot[i]+=ma[j][k];
				}
				break;
			}
		}
	}
	memset(dp,-0x3f3f3f,sizeof(dp));
	dp[1][1]=0;
	for(int i=1;i<(1<<n);++i)
	{
		for(int j=1;j<=n;++j)
		{
			if(i&(1<<(j-1)))
			{
				int r=i;
				while(r)
				{
					if(r&(1<<(j-1)))
					{
						getmax(dp[i][j],dp[(i-r)|(1<<(j-1))][j]+tot[r]);
					}
					r=(r-1)&i;
				}
				for(int k=1;k<=n;++k)
				{
					if(!(i&(1<<(k-1))))
					{
						getmax(dp[i|(1<<(k-1))][k],dp[i][j]+ma[j][k]);
					}
				}
			}
		}
	}
	int res=0;
	for(int i=0;i<(1<<n);++i)
	{
		if(i&(1<<(n-1)))
		getmax(res,dp[i][n]);
	}
	cout<<tt-res<<endl;
  return 0;
}