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;
typedef long long ll;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
ll n,m,g[15][15],a[1<<15],dp[15][1<<15];
int main(){
	n=read(),m=read();
	for(int i=0;i<m;i++){
		int x=read()-1,y=read()-1,z=read();
		g[x][y]=g[y][x]=z;
	}
	for(int i=0;i<1<<n;i++){
		a[i]=0;
		for(int j=0;j<n;j++)
			for(int k=j+1;k<n;k++)
				if(i&(1<<j) and i&(1<<k))
					a[i]+=g[j][k];
	}
	memset(dp,-0x3f,sizeof(dp));
	for(int i=1;i<1<<n;i+=2)
		dp[0][i]=a[i];
	for(int i=0;i<1<<n;i++)
		for(int j=0;j<n;j++){
			if(dp[j][i]<0)continue;
			ll ab=(1<<n)-1-i;
			for(int p=ab;p>0;p=(p-1)&ab)
				for(int k=0;k<n;k++)
					if(g[j][k] and(p&(1<<k)))
						dp[k][i+p]=max(dp[k][i+p],dp[j][i]+a[p]+g[j][k]);
		}
	ll ans=0;
	for(int i=0;i<n;i++)
		for(int j=i+1;j<n;j++)
			ans+=g[i][j];
	printf("%lld",ans-dp[n-1][(1<<n)-1]);
	return 0;
}