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<cstdio> 
#include<algorithm>
#include<cstring>
#include<cmath>
#include<map>
using namespace std;
#define fo(i,j,k) for(i=j;i<=k;i++)
#define fd(i,j,k) for(i=j;i>=k;i--)
#define cmax(a,b) (a=(a>b)?a:b)
#define cmin(a,b) (a=(a<b)?a:b)
typedef long long ll;
typedef double db;
const int N=16,M=1<<16;
int f[M][N],g[M],sum[M],i,j,s,ts,rs,s1,k,n,m,x,y,z,p,mx,tot;
int tt,b[N*N],c[N*N],nxt[N*N],fst[N];
void cr(int x,int y,int z)
{
	tt++;
	b[tt]=y;
	c[tt]=z;
	nxt[tt]=fst[x];
	fst[x]=tt;
}
int main()
{
	scanf("%d %d",&n,&m);
	fo(i,1,m)
	{
		scanf("%d %d %d",&x,&y,&z);
		tot+=z;
		cr(x,y,z);
		cr(y,x,z);
	}
	ts=1<<n;ts--;
	fo(i,0,ts) fo(j,1,n) f[i][j]=-1;
	f[1][1]=0;
	fo(i,0,ts)
		fo(j,1,n)
		if (f[i][j]!=-1)
		{
			for(p=fst[j];p;p=nxt[p])
			{
				k=b[p];
				if (!((1<<k-1)&i))
					cmax(f[i+(1<<k-1)][k],f[i][j]+c[p]);
			}
		}
	fo(i,0,ts) g[i]=f[i][n];
	fo(i,0,ts)
	{
		fo(j,1,n) if (i&(1<<j-1))
		{
			for(p=fst[j];p;p=nxt[p])
			{
				k=b[p];
				if ((i&(1<<k-1))&&k<j) 
					sum[i]+=c[p];
			}
		}
	}
	fo(s,0,ts)
	if (g[s]!=-1)
	{
		rs=s^ts;
		for(s1=rs;s1;s1=(s1-1)&rs)
		{
			mx=0;
			fo(j,1,n) if (s&(1<<j-1))
				cmax(mx,sum[s1+(1<<j-1)]);
			cmax(g[s+s1],g[s]+mx);
		}
	}
	printf("%d\n",tot-g[ts]);
}