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;
template<typename T>bool chkmax(T &a,T b){
	if(a<b){a=b;return 1;}
	else return 0;
}
template<typename T>T read(){
	T x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-')f=-1;
	for(;isdigit(c);c=getchar())x=(x<<3)+(x<<1)+(c^48);
	return x*f;
}
const int maxn=1<<15;
int n,m,S,sum,ans=2e9,c[15][15],p[maxn],dp[maxn][15];
void init(){
	memset(dp,-1,sizeof dp);
	S=1<<n;
	for(int s=0;s<S;++s)
		for(int i=0;i<n;++i){
			if(((s>>i)&1)==0)continue;
			for(int j=i+1;j<n;++j)
				if((s>>j)&1)p[s]+=c[i][j];
		}
	dp[1][0]=0;
}
void solve(){
	n=read<int>(),m=read<int>();
	for(int u,v,w;m--;){
		u=read<int>()-1;
		v=read<int>()-1;
		w=read<int>();
		c[u][v]=c[v][u]=w;
		sum+=w;
	}
	init();
	for(int s=0;s<S;++s){
		for(int i=0;i<n;++i){
			if(!~dp[s][i])continue;
			int x=((S-1)^s)|(1<<i);
			for(int t=x;~t;--t)
				t&=x,chkmax(dp[s|t][i],dp[s][i]+p[t]);	
			for(int j=0;j<n;++j)if((~(s>>j)&1)&&c[i][j]){
				chkmax(dp[s|(1<<j)][j],dp[s][i]+c[i][j]);
			}
		}
	}
	for(int i=0;i<S;++i)
		ans=min(ans,sum-dp[i][n-1]);
	printf("%d\n",ans);
}
int main(){
	solve();
	
	return 0;
}