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<iostream>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<fstream>
#include<map>
#include<queue>
#include<stack>
using namespace std;
long long n,m,sum,c[20][20];
long long f[1<<15][15],p[1<<15];
int main(){
	memset(f,-1,sizeof(f));
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		int u,v,w;
		cin>>u>>v>>w;
		u--,v--;
		sum+=w;
		c[u][v]=w,c[v][u]=w;
	}
	for(int i=0;i<=(1<<n)-1;i++){
		for(int j=0;j<=n-1;j++){
			if(i&(1<<j)){
				for(int k=j+1;k<=n-1;k++){
					if(i&(1<<k)){
						p[i]=p[i]+c[j][k];
					}
				}
			}
		}
	}
	f[1][0]=0;
	for(int i=0;i<=(1<<n)-1;i++){
		for(int j=0;j<=n-1;j++){
			if(f[i][j]==-1){
				continue;
			}
			int x=(((1<<n)-1)^i)|(1<<j);
			for(int k=x;k;k=(k-1)&x){
				f[i|k][j]=max(f[i|k][j],f[i][j]+p[k]);
			}
			for(int k=1;k<=n-1;k++){
				if((!(i&(1<<k)))&&c[j][k]){
					f[i|(1<<k)][k]=max(f[i|(1<<k)][k],f[i][j]+c[j][k]);
				}
			}
		}
	}
	long long ans=100000000000000;
	for(int i=0;i<=(1<<n)-1;i++){
		for(int j=0;j<=n-1;j++){
			if(i&(1<<j)){
				ans=min(ans,sum-f[i][n-1]);
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}