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 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)
#define fill(x,t) memset(x,t,sizeof(x))
#define lb(x) (x&-x)
#define get(x) ((int)log2(x))

typedef long long LL;
const int N=200005;

LL r[20][N],f[20][N],sum[N],g[N];
LL v[20][20];

int read() {
	int x=0,v=1; char ch=getchar();
	for (;ch<'0'||ch>'9';v=(ch=='-')?(-1):v,ch=getchar());
	for (;ch<='9'&&ch>='0';x=x*10+ch-'0',ch=getchar());
	return x*v;
}

void upd(LL &x,LL v) {
	(x<v)?(x=v):0;
}

int main(void) {
	int n=read(),m=read();
	LL prt=0;
	rep(i,1,m) {
		int x=read()-1,y=read()-1;
		v[x][y]=v[y][x]=read(),prt+=v[x][y];
		r[x][1<<y]=v[x][y];
		r[y][1<<x]=v[x][y];
	}
	rep(i,0,n-1) {
		for (int S=0;S<(1<<n);++S) {
			r[i][S]=r[i][S-lb(S)]+r[i][lb(S)];
		}
	}
	for (int S=0;S<(1<<n);++S) {
		sum[S]=sum[S-lb(S)]+r[get(lb(S))][S-lb(S)];
	}
	fill(f,-63); f[0][1]=0;
	LL ans=0;
	rep(i,0,n-1) {
		for (int S=(1<<n)-1;S;--S) {
			if (!((S>>i))&1) continue;
			if (!(S&1)) continue;
			for (int T=S;T;T=(T-1)&S) {
				if (T>>i&1) upd(f[i][S],f[i][S-T+(1<<i)]+sum[T]);
			}
			rep(j,0,n-1) if (!(S>>j&1)&&v[i][j]) {
				upd(f[j][S+(1<<j)],f[i][S]+v[i][j]);
			}
		}
	}
	for (int S=0;S<(1<<n);++S) if ((S&1)&&(S>>(n-1)&1)) upd(ans,f[n-1][S]);
	printf("%lld\n", prt-ans);
	return 0;
}