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<algorithm>
using namespace std;
const int N=16,M=N*N,K=1<<15,inf=1e9;
int n,m,nm,i,j,k,x,p[K],v[K],f[N][K],mx[N];
int et,he[N],sum;
struct edge{int l,to,v;}e[M];
void gmax(int &x,int k){if (k>x) x=k;}
int main(){
	scanf("%d%d",&n,&m);nm=1<<n;
	for (i=1;i<=m;i++){
		int x,y,v;scanf("%d%d%d",&x,&y,&v);sum+=v;
		e[++et].l=he[x];he[x]=et;e[et].to=y;e[et].v=v;
		if (x==y) continue;
		e[++et].l=he[y];he[y]=et;e[et].to=x;e[et].v=v;
	}
	for (i=1;i<=n;i++) p[1<<i-1]=i;
	for (i=1;i<nm;i++){
		int x=p[i&-i];v[i]=v[i-(i&-i)];
		for (j=he[x];j;j=e[j].l)
			if ((i>>e[j].to-1)&1) v[i]+=e[j].v;
	}
	for (x=1;x<=n;x++)
		for (i=0;i<nm;i++) f[x][i]=-inf;
	for (i=1;i<nm;i+=2) f[1][i]=v[i];
	for (i=1;i<nm-1;i++) if (i&1){
		for (x=1;x<=n;x++) mx[x]=-inf;
		for (x=1;x<=n;x++){
			if (f[x][i]==-inf) continue;
			for (j=he[x];j;j=e[j].l)
				gmax(mx[e[j].to],f[x][i]+e[j].v);
		}
		for (j=i;j<nm;j=(j+1)|i)
			for (k=j^i;k;k-=k&-k){
				int x=p[k&-k];
				gmax(f[x][j],mx[x]+v[j^i]);
			}
	}
	printf("%d",sum-f[n][nm-1]);
}