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 <cstdio>
#include <cstring>
#define MAXN 20
#define MAXS 40000

int n,m,maxs;
int g[MAXS],f[MAXS][MAXN];
int e[MAXN][MAXN];

inline void update(int &x,int y){ x=x>y?x:y; }

int main(){
#ifdef DEBUG
	freopen("F.in","r",stdin);
#endif
	scanf("%d%d",&n,&m);
	int totw=0;
	for(int i=1;i<=m;i++){
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		e[v][u]=e[u][v]=w;
		totw+=w;
	}
	maxs=1<<n;
	for(int i=0;i<maxs;i++){
		for(int j=1;j<=n;j++)
			if(i&(1<<(j-1)))
				for(int k=j+1;k<=n;k++)
					if(i&(1<<(k-1)))
						g[i]+=e[j][k];
	}
	int maxs0=maxs;
	maxs>>=2;
	memset(f,0x88,sizeof f);
	for(int i=0;i<maxs;i++) f[i][1]=g[i<<1|1];
	for(int i=0;i<maxs;i++)
		for(int x=1;x<n;x++){
			if(x!=1 && !(i&(1<<(x-2)))) continue;
			for(int y=2;y<n;y++){
				if((i&(1<<(y-2))) || !e[x][y]) continue;
				int by=1<<(y-2);
				int lims=(maxs-1)^i^by;
				for(int j=lims;;j=(j-1)&lims){
					update(f[i|j|by][y],f[i][x]+e[x][y]+g[(j|by)<<1]);
					if(!j) break;
				}
			}
		}
	int ans=-1;
	for(int i=0;i<maxs;i++)
		for(int j=1;j<n;j++)
			if(j==1 || (i&(1<<(j-2)))){
				if(e[j][n])
					update(ans,f[i][j]+e[j][n]+g[(maxs0-1)^1^(i<<1)]);
			}
	ans=totw-ans;
	if(ans<=totw) printf("%d\n",ans);
	else puts("-233");
}