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>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

inline int Read()
{
	int x = 0, f = 1, c = getchar();
	for (; !isdigit(c); c = getchar())
		if (c == '-')
			f = -1;
	for (;  isdigit(c); c = getchar())
		x = x * 10 + c - '0';
	return x * f;
}

const int MAXN = 20;
const int MAXM = 32780;

int n, m, tot, val[MAXM], adj[MAXN][MAXN], f[MAXM][MAXN];

int main()
{
#ifdef wxh010910
	freopen("data.in", "r", stdin);
#endif
	n = Read(), m = Read();
	for (int i = 1, x, y, w; i <= m; i ++)
		x = Read() - 1, y = Read() - 1, w = Read(), adj[x][y] = adj[y][x] = w, tot += w;
	for (int i = 0; i < 1 << n; i ++)
		for (int j = 0; j < n; j ++)
			if (i >> j & 1)
				for (int k = j + 1; k < n; k ++)
					if (i >> k & 1)
						val[i] += adj[j][k];
	mset(f, -1);
	for (int i = 1; i < 1 << n; i += 2)
		f[i][0] = val[i];
	for (int i = 1; i < 1 << n; i += 2)
		for (int j = 0; j < n; j ++)
			if (~f[i][j])
			{
				for (int s = (1 << n) - 1 ^ i, k = s; k; k = k - 1 & s)
					f[i | k][j] = max(f[i | k][j], f[i][j] + val[k | 1 << j]);
				for (int k = 0; k < n; k ++)
					if (adj[j][k] && !(i >> k & 1))
						f[i | 1 << k][k] = max(f[i | 1 << k][k], f[i][j] + adj[j][k]);
			}
	return printf("%d\n", tot - f[(1 << n) - 1][n - 1]), 0;
}