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 I[40000],n,m,H[20],a[20][20],x,y,ans,f[20][35000],w[35000];
int main(){
	scanf("%d%d",&n,&m);
	I[1]=H[1]=1; for (int i=2;i<=n+1;++i) H[i]=H[i-1]<<1,I[H[i]]=i;
	for (int i=1;i<=m;++i){
		scanf("%d%d",&x,&y);
		scanf("%d",&a[x][y]);
		a[y][x]=a[x][y];
		for (int j=1;j<H[n+1];++j)
		if (j&H[x]&&j&H[y]) w[j]+=a[x][y];
	}
	for (int i=1;i<=n;++i)
	for (int j=0;j<H[n+1];++j) f[i][j]=-1;
	for (int s=1;s<H[n];s+=2) f[1][s]=w[s];
	for (int s=1;s<H[n];s+=2){
		for (int i=2;i<n;++i)
		if (H[i]&s){
			int S=s^H[i];
			for (int o=S;o;o=o-1&S)
			if (o&1){
				for (int j=o;j;j-=j&-j)
				if (a[i][I[j&-j]])
					f[i][s]=max(f[i][s],a[i][I[j&-j]]+f[I[j&-j]][o]+w[s-o]);
			}
		}
	}
	for (int i=1;i<n;++i)
	if (a[i][n])
	for (int s=1;s<H[n];s+=2)
	if (f[i][s]>-1)
	ans=max(ans,f[i][s]+a[i][n]+w[H[n+1]-1-s]);
	printf("%d\n",w[H[n+1]-1]-ans);
	return 0;
}