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:
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
using namespace std;
#define chmax(a,b) if(a<b)a=b
const int N=17;

int n,m;
int g[N][N];
int cost[1<<N];
int dp[N][1<<N],dp2[1<<N];

int main()
{
	
	scanf("%d%d",&n,&m);
	while(m--)
	{
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		u--; v--;
		g[u][v]=g[v][u]=w;
	}
	for(int mask=0;mask<(1<<n);mask++)
	for(int u=0;u<n;u++)
	for(int v=u+1;v<n;v++)
		if(mask&(1<<u) && mask&(1<<v)) cost[mask]+=g[u][v];
	memset(dp,-1,sizeof(dp));
	for(int mask=1;mask<(1<<n);mask+=2)
		dp[0][mask]=cost[mask];
	for(int mask=1;mask<(1<<(n-1));mask+=2)
	for(int i=0;i<n-1;i++) if(dp[i][mask]!=-1)
	for(int j=0;j<n;j++)
		if(g[i][j] && !(mask&(1<<j))) chmax(dp[j][mask|(1<<j)],dp[i][mask]+g[i][j]);
	memset(dp2,-1,sizeof(dp2));
	for(int mask=1;mask<(1<<n);mask+=2)
		dp2[mask]=dp[n-1][mask];
	for(int mask=1+(1<<(n-1));mask<(1<<n);mask+=2) if(dp2[mask]!=-1)
	for(int i=0;i<n;i++) if(mask&(1<<i))
	{
		int _=(1<<n)-1-mask;
		for(int mask2=_;mask2>0;mask2=(mask2-1)&_)
			chmax(dp2[mask|mask2],dp2[mask]+cost[mask2|(1<<i)]);
	}
	printf("%d\n",cost[(1<<n)-1]-dp2[(1<<n)-1]);
	
	return 0;
}