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++ (GCC 9.2.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast")
#include<stdio.h>
#include<algorithm>
using namespace std;
int g[20][20]={0},cost[1<<15]={0},dp[1<<15][15];
int main(){
	int i,j,k,n,m,a,b,c;
	scanf("%d%d",&n,&m);
	for(i=0;i<m;i++){
		scanf("%d%d%d",&a,&b,&c);
		g[a-1][b-1]=g[b-1][a-1]=c;
	}
	for(i=1;i<(1<<n);i++){
		for(j=0;j<n;j++){
			if(!(i>>j&1))
				continue;
			for(k=j+1;k<n;k++)
				if(i>>k&1)
					cost[i]+=g[j][k];
		}
	}
	for(i=0;i<(1<<n);i++)
		for(j=0;j<n;j++)
			dp[i][j]=-1000000000;
	for(i=0;i<(1<<n);i++)
		dp[i][0]=cost[i];
	for(i=0;i<(1<<n);i++){
		for(j=1;j<n;j++){
			if(!(i>>j&1))
				continue;
			for(k=0;k<n;k++)
				if(k!=j&&(i>>k&1)&&g[j][k])
					dp[i][j]=max(dp[i][j],dp[i^(1<<j)][k]+g[j][k]);
			for(k=i;k;k=(k-1)&i)
				dp[i][j]=max(dp[i][j],dp[i^k][j]+cost[k|(1<<j)]);
		}
	}
	printf("%d\n",cost[(1<<n)-1]-dp[(1<<n)-1][n-1]);
}