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<cstdlib>
#include<cstring>
#include<algorithm>
#define inf 0x3f3f3f3f
#define MAX(x,y) x=max(x,y)
using namespace std;
 
const int N=18;
const int M=32775;
 
int n,m,bin[N],num[M],ma[N][M],sum[M],f[N][M],g[M];
 
int read()
{
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
 
int lowbit(int x)
{
	return x&(-x);
}
 
int main()
{
	n=read();m=read();
	bin[0]=1;
	for (int i=1;i<=n;i++) bin[i]=bin[i-1]*2,num[bin[i]]=i;
	int tot=0;
	for (int i=1;i<=m;i++)
	{
		int x=read()-1,y=read()-1,z=read();tot+=z;
		ma[x][bin[y]]=ma[y][bin[x]]=z;
	}
	for (int i=0;i<n;i++)
		for (int j=1;j<bin[n];j++)
			ma[i][j]=ma[i][j-lowbit(j)]+ma[i][lowbit(j)];
	for (int i=1;i<bin[n];i++) sum[i]=sum[i-lowbit(i)]+ma[num[lowbit(i)]][i-lowbit(i)];
	for (int i=0;i<n;i++)
		for (int j=0;j<bin[n];j++)
			f[i][j]=-inf;
	f[0][1]=0;
	for (int j=0;j<bin[n];j++)
		for (int i=0;i<n-1;i++)
		{
			if (f[i][j]<0) continue;
			for (int k=0;k<n;k++)
				if (!(j&bin[k])&&ma[i][bin[k]]) MAX(f[k][j+bin[k]],f[i][j]+ma[i][bin[k]]);
		}
	for (int i=0;i<bin[n];i++) g[i]=f[n-1][i];
	for (int i=1;i<bin[n];i++)
	{
		if (g[i]<0) continue;
		int r=i^(bin[n]-1);
		for (int s=r;s;s=(s-1)&r)
		{
			int mxv=0;
			for (int j=i;j;j-=lowbit(j)) mxv=max(mxv,sum[s+lowbit(j)]);
			MAX(g[i+s],g[i]+mxv);
		}
	}
	printf("%d",tot-g[bin[n]-1]);
	return 0;
}