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 g[15][15],cr[15][32768],ins[32768],dp[32768][15];
int main (){
	int i,j,k,t,u,v,c,l,n,m,s=0;
	scanf ("%d%d",&n,&m);
	for (i=1;i<=m;i++)
	{scanf ("%d%d%d",&u,&v,&c);
	u--;v--;g[u][v]=g[v][u]=c;s+=c;
	}
	for (i=0;i<n;i++)
	{for (k=0;k<n;k++)
	{for (j=0;j<(1<<n);j++)
	{if ((1<<k)&j) 
	{cr[i][j]+=g[i][k];}
	}
	}
	}
	for (i=0;i<(1<<n);i++)
	{for (j=0;j<n;j++)
	{for (k=0;k<n;k++)
	{if ((1<<j)&i)
	{if ((1<<k)&i)
	{ins[i]+=g[j][k];}
	}
	}
	}
	ins[i]/=2;
	}
	memset(dp,-1,sizeof(dp));
	dp[1][0]=0;
	for (i=1;i<(1<<n);i++)
	{for (j=0;j<n;j++)
	{if (dp[i][j]<0) {continue;}
	int tar=i-(1<<j),b=(1<<n)-1-i;
	for (k=0;k<n;k++)
	{if (i&(1<<k)) {continue;}
	if (g[j][k]<=0) {continue;}
	dp[i+(1<<k)][k]=max(dp[i+(1<<k)][k],dp[i][j]+g[j][k]);
	}
	for (t=b;t;t=(t-1)&b)
	{dp[t|i][j]=max(dp[t|i][j],dp[i][j]+cr[j][t]+ins[t]);}
	}
	}
	int ans=0;
	for (i=0;i<(1<<n);i++)
	{ans=max(ans,dp[i][n-1]);}
	printf ("%d",s-ans);
	return 0;
}