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 std::max;
 
const int N=15,Inf=1e9+10;
 
int n,m,map[N][N],f[N][1<<N],M[N][1<<N],sum[1<<N];
 
void Init() {
	scanf("%d%d",&n,&m);
	int a,b,c;
	for (int i=1;i<=m;++i) {
		scanf("%d%d%d",&a,&b,&c);
		a--; b--; 
		map[a][b]=map[b][a]=c;
	}
}
 
int summ;
bool vis[N];
void Dfs(int x,int s) {
	if (vis[x]) return ;
	vis[x]=1;
	for (int i=0;i<n;++i) if ((1<<i)&s && map[x][i]) {
		summ+=map[x][i];
		Dfs(i,s);
	}
}
 
void Solve() {
	int Max=1<<n;
	for (int s=0;s<Max;++s) {
		summ=0;
		for (int i=0;i<n;++i) vis[i]=0;
		for (int i=0;i<n;++i) if ((1<<i)&s) Dfs(i,s);
		sum[s]=summ>>1;
	}
	
	summ=0;
	for (int i=0;i<n;++i)
		for (int j=i;j<n;++j) summ+=map[i][j];
	for (int i=0;i<n;++i)
		for (int s=0;s<Max;++s) f[i][s]=-Inf,M[i][s]=-Inf;
	M[0][1]=0;
	for (int s=0;s<Max;++s) if (s&1) {
		f[0][s]=sum[s];
		for (int t=0;t<n;++t) 
			if (map[t][0]) M[t][s]=max(M[t][s],f[0][s]+map[t][0]);
	}
	for (int s=0;s<Max;++s) {
		for (int i=0;i<n;++i) if ((1<<i)&s) {
			for (int p=s;p;p=(p-1)&s) if (!((1<<i)&p)) {
				f[i][s]=max(f[i][s],M[i][p]+sum[s^p]);
			}
			for (int t=0;t<n;++t) if (map[t][i]) M[t][s]=max(M[t][s],f[i][s]+map[t][i]);
		}
	}
	
	printf("%d\n",summ-f[n-1][Max-1]);
}
 
int main() {
	Init();
	Solve();
	return 0;
}