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 <cstring>
using namespace std;

const int maxn=17, maxst=(1<<17)-1;
int n, m, sum, e[maxn][maxn], p[maxst], f[maxst][maxn];

inline int max(int x, int y){ return x<y?y:x; }
inline void up(int &x, int y){ x=max(x, y); }

int main(){
	scanf("%d%d", &n, &m); int a, b, c, ans=0;
	memset(f, -1, sizeof(f));  //还是那个，不能让不合法状态成功转移 
	for (int i=0; i<m; ++i){
		scanf("%d%d%d", &a, &b, &c); --a; --b;
		e[a][b]=e[b][a]=c; sum+=c; }
	for (int i=0; i<1<<n; ++i){
		for (int j=0; j<n; ++j) if (i&(1<<j))
		for (int k=0; k<n; ++k) if (i&(1<<k))
			p[i]+=e[j][k];
		p[i]>>=1; }
	f[1][0]=0;
	for (int i=1; i<(1<<n); ++i){  //当前集合 
		for (int j=0; j<n; ++j){  //当前挂的点 
			if (f[i][j]==-1) continue; 
			for (int k=0; k<n; ++k){  //枚举新在j后加的点
				if (i&(1<<k)) continue;
				if (e[j][k]==0) continue;  //绝壁想不到的东西！果然转移要想清楚，限制一定要先治好，不能让不合法状态成功转移 
				up(f[i|(1<<k)][k], f[i][j]+e[j][k]);
			}
			int revi=(((1<<n)-1)^i)|(1<<j);  //枚举新挂在j上的集合（要加上j以统计连j的边） 
			for (int k=revi; k; k=(k-1)&revi)
				up(f[i|k][j], f[i][j]+p[k]);
		}
	}
	for (int i=0; i<(1<<n); ++i)
		up(ans, f[i][n-1]);
	printf("%d\n", sum-ans);
	return 0;
}