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>
#define N 18
using namespace std;
int dis[N][N];
int n,m;
int tot[(1<<15)+3];
int dp[(1<<15)+3][N];
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1,f,t,d;i<=m;i++){
		scanf("%d%d%d",&f,&t,&d);
		dis[f-1][t-1]=dis[t-1][f-1]=d;
	}
	
	for(int i=1;i<(1<<n);i++){
		int pos=-1,aim=i-(i&(-i));
		for(int j=0;j<n;j++)
			if((i&(-i))>>j == 1){
				pos=j;break;
			}
		if(pos!=-1){
			tot[i]=tot[aim];
			for(int j=0;j<n;j++)
				if(aim&(1<<j)) tot[i]+=dis[pos][j];
		}
	}
	
	memset(dp,0xc0,sizeof(dp));
	for(int i=1;i<(1<<n);i+=2)
		dp[i][0]=tot[i];
	for(int i=1;i<(1<<n);i++)
		for(int j=0;j<n;j++)
			if(i&(1<<j)&&dp[i][j]!=dp[0][0]){
				for(int k=0;k<n;k++)
					if((i&(1<<k)) == 0&&dis[j][k] != 0){
						for(int mask=((1<<n)-1)^(i|(1<<k)),v=mask;;v=(v-1)&mask){
							dp[i|(1<<k)|v][k]=max(dp[i|(1<<k)|v][k],dp[i][j]+dis[j][k]+tot[v|(1<<k)]);
							if(v==0) break;
						}
					}
			}
		
	int ans=-1e9;
	for(int i=(1<<(n-1));i<(1<<n);i++) ans=max(ans,dp[i][n-1]);
	cout<<tot[(1<<n)-1]-ans<<endl;
	return 0;
}