Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

4 6
1 2 100
3 1 100
2 4 100
4 3 100
1 4 100
3 2 100

Sample Output 1

200

By removing the two edges represented by the red dotted lines in the figure below, the objective can be achieved for a cost of 200 yen.

Sample Input 2

2 1
1 2 1

Sample Output 2

0

It is possible that there is already only one path from Vertex 1 to Vertex N in the beginning.

Sample Input 3

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

Sample Output 3

133677

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define maxn 40010
using namespace std;
int p[maxn],dp[maxn][15];
int c[15][15];
int n,m;
void init(){//printf("&\n");
		for(int i=0;i<(1<<n);i++)for(int j=0;j<n;j++)dp[i][j]=-1;
		for(int i=0;i<(1<<n);i++){
				for(int j=0;j<n;j++){
						if((i>>j)&1){
								for(int k=j+1;k<n;k++){
										if((i>>k)&1){
												p[i]+=c[j][k];
										}
								}
						}
				}
		}
		//	printf("&&\n");
}
int main(){
		while(~scanf("%d%d",&n,&m)){
				//scanf("%d%d",&n,&m);
				int sum=0;
				memset(c,0,sizeof c);
				while(m--){
						int u,v,w;scanf("%d%d%d",&u,&v,&w);
						u--;v--;sum+=w;
						c[u][v]=c[v][u]=w;
				}
				memset(p,0,sizeof p);init();
				dp[1][0]=0;
				for(int i=0;i<(1<<n);i++){
						for(int j=0;j<n;j++){
								if(dp[i][j]==-1)continue;
								int x=(((1<<n)-1)^i)|(1<<j);
								for(int y=x;y>=0;y--){
										y&=x;
										dp[i|y][j]=max(dp[i|y][j],dp[i][j]+p[y]);
								}
								for(int k=0;k<n;k++){
										if(!c[j][k])continue;
										if((i>>k)&1)continue;
										dp[i|(1<<k)][k]=max(dp[i|(1<<k)][k],dp[i][j]+c[j][k]);
								}//printf("&\n");
						}
				}//printf("*");
				int ans=1e9+7;
				for(int i=0;i<(1<<n);i++){
						ans=min(ans,sum-dp[i][n-1]);
				}
				printf("%d\n",ans);
		}
		return 0;
}