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 <bits/stdc++.h>
int n,m,x,y,z;
int a[20][20],sum,dp[1<<15][15],f[1<<15];
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++){
		scanf("%d%d%d",&x,&y,&z);
		x--,y--;
		a[x][y]=a[y][x]=z;
		sum+=z;
	}
	for (int i=1;i<1<<n;i++)
		for (int j=0;j<n;j++)
			for (int k=j+1;k<n;k++)
				if ((i&(1<<j))&&(i&(1<<k)))
					f[i]+=a[j][k];
	memset(dp,-1,sizeof(dp));
	dp[1][0]=0;
	int all=(1<<n)-1;
	for (int i=1;i<1<<n;i+=2){
		for (int j=0;j<n;j++)
			if (i&(1<<j)){
				int t=dp[i][j];
				if (t==-1) continue;
				for (int k=0;k<n;k++){
					if (i&(1<<k)) continue;
					if (a[j][k]) dp[i|(1<<k)][k]=std::max(dp[i|(1<<k)][k],t+a[j][k]);
				}
				int rest=all^i;
				for (int k=rest;k;k=(k-1)&rest){
					dp[i|k][j]=std::max(dp[i|k][j],t+f[k|(1<<j)]);
				}
			}
	} 
	printf("%d\n",sum-dp[all][n-1]);
}