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<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const int INF=~0u>>2;
int n,m,sum[1<<15],dp[1<<15][15],G[15][15];
int read(){
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return f*x;
}
int cost(int x,int y){return sum[x|y]-sum[x]-sum[y];}
int main(){
//	freopen("ex.in","r",stdin);
	n=read();m=read();
	for(int i=1;i<=m;++i){
		int x=read(),y=read(),z=read();
		x--;y--;
		G[x][y]=G[y][x]=z;
	}
	for(int i=1;i<(1<<n);++i){
		for(int j=0;j<n;++j) if(i&(1<<j)){
			for(int k=j+1;k<n;++k) if(i&(1<<k)) sum[i]+=G[j][k];
		}
	}
	for(int i=1;i<(1<<n);++i) for(int j=0;j<n;++j) dp[i][j]=INF;
	dp[1][0]=0;
	for(int i=1;i<(1<<n);++i){
		for(int j=0;j<n;++j) if(dp[i][j]!=INF){
			for(int k=0;k<n;++k) if(!(i&(1<<k))&&G[j][k]) dp[i|(1<<k)][k]=min(dp[i|(1<<k)][k],dp[i][j]+cost(i^(1<<j),1<<k));
			for(int k=((1<<n)-1)&(~i);k>0;k=(k-1)&(~i)) dp[i|k][j]=min(dp[i|k][j],dp[i][j]+cost(i^(1<<j),k));
		}
	}
	printf("%d",dp[(1<<n)-1][n-1]);
	return 0;
}