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,m,edge[20][20];
long long ans=0,sum[1<<16],dp[1<<16][20];
int main(){
	scanf("%d %d",&n,&m);
	for(int i=1,u,v,w;i<=m;i++){
		scanf("%d %d %d",&u,&v,&w);
		edge[u][v]=edge[v][u]=w;
		ans+=w;
	}
	int tmp=(1<<n);
	for(int S=0;S<tmp;S++){
		for(int i=1;i<=n;i++){
			dp[S][i]=-1;
			if(!(S&(1<<(i-1)))&&!sum[S|(1<<(i-1))]){
				sum[S|(1<<(i-1))]=sum[S];
				for(int j=1;j<=n;j++){
					if(S&(1<<(j-1))) sum[S|(1<<(i-1))]+=edge[i][j];
				}
			}
		}
	}
	dp[1][1]=0;
	for(int S=0;S<tmp;S++){
		for(int i=1;i<=n;i++){
			if(dp[S][i]<0) continue;
			if(S&(1<<(i-1))){
				for(int j=1;j<=n;j++){
					if(!(S&(1<<(j-1)))&&edge[i][j]){
						dp[S|(1<<(j-1))][j]=max(dp[S][i]+edge[i][j],dp[S|(1<<(j-1))][j]);
					}
				}
				int now=((tmp-1)^S)|(1<<(i-1));
				for(int M=now;M;M=(M-1)&now){
					if(M&(1<<(i-1))) dp[S|M][i]=max(dp[S][i]+sum[M],dp[S|M][i]);
				}
			}
		}
	}
	printf("%lld\n",ans-dp[tmp-1][n]);
}