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;
const int Max_N=15,Max_S=1<<15;
int N,M,Sum=0,A[Max_N+5][Max_N+5]={},f[Max_S+5][Max_N+5]={},P[Max_S+5]={};
int main(){
	scanf("%d%d",&N,&M);
	for(int i=1;i<=M;i++){
		int x,y,v; 
		scanf("%d%d%d",&x,&y,&v);
		Sum+=v; A[x][y]=A[y][x]=v;
	}
	for(int i=0;i<1<<N;i++){
		for(int j=1;j<=N;j++)
		 if((1<<(j-1))&i){
		 	for(int k=j+1;k<=N;k++)
		 	 if((1<<(k-1))&i)
		 	  P[i]+=A[j][k];
		 }
	}
	for(int i=1;i<=N;i++)
	 for(int j=0;j<1<<N;j++)
	  f[j][i]=-(1<<30);
	f[1][1]=0;
	for(int i=0;i<1<<N;i++){
		for(int j=1;j<=N;j++){
			if((1<<(j-1))&i)
			for(int k=1;k<=N;k++)
			 if(((1<<(k-1))&i)&&k!=j&&A[j][k]>0)
			  f[i][j]=max(f[i][j],f[i^(1<<(j-1))][k]+A[j][k]);
			for(int k=i;k;k=(k-1)&i)
			 if(k&(1<<(j-1)))
			  f[i][j]=max(f[i][j],f[(i^k)|(1<<(j-1))][j]+P[k]);
		}
	}
	printf("%d\n",Sum-f[(1<<N)-1][N]);
	return 0;
}