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;
const int N=20,M=200000;
int a[N][N],f[M][N],g[M][N],n,m,x,y,z;
inline void read(int &x)
{
	char c=getchar();
	x=0;
	while (c>'9'||c<'0')
		c=getchar();
	while (c>='0'&&c<='9')
	{
		x=x*10+c-'0';
		c=getchar();
	}
}
inline int work(int x,int y)
{
	int sum=0;
	for (int i=0;i<n;++i)
		if ((x>>i)&1)
			sum+=f[y][i];
	return sum;
}
int main()
{
	read(n);
	read(m);
	memset(g,0x3f,sizeof(g));
	for (int i=1;i<=m;++i)
	{
		read(x);
		read(y);
		read(z);
		--x;
		--y;
		a[x][y]=a[y][x]=z;
	}
	for (int i=0;i<(1<<n);++i)
		for (int j=0;j<n;++j)
			for (int k=0;k<n;++k)
				if ((i>>k)&1)
					f[i][j]+=a[j][k];
	g[1][0]=0;
	for (int i=0;i<(1<<n);++i)
		for (int j=0;j<n;++j)
			if ((i>>j)&1)
			{
				for (int k=0;k<n;++k)
					if (a[j][k]&&(!((i>>k)&1)))
						g[i|(1<<k)][k]=min(g[i|(1<<k)][k],g[i][j]+work(i,1<<k)-a[j][k]);
				int u=(1<<n)-1-i;
				for (int k=u;k;k=(k-1)&u)
					g[i|k][j]=min(g[i|k][j],g[i][j]+work(i^(1<<j),k));
			}
	printf("%d",g[(1<<n)-1][n-1]);
	return 0;
}