Problem: 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.

Examples

Input

4 6
1 2 100
3 1 100
2 4 100
4 3 100
1 4 100
3 2 100


Output

200


Input

2 1
1 2 1


Output

0


Input

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


Output

133677

Solution (in CPP):

Solution:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
int n, m, map[16][16];
ll in[1 << 16], f[1 << 16][16];
int main()
{
	scanf("%d%d", &n, &m);
	int x, y;
	for (int i = 1; i <= m; i++)
	{
		scanf("%d%d", &x, &y);
		x--, y--;
		scanf("%d", &map[x][y]);
		map[y][x] = map[x][y];
	}
	for (int i = 0; i < (1 << n); i++)
		for (int j = 0; j < n; j++)
			if (i & (1 << j))
				for (int k = j + 1; k < n; k++)
					if (i & (1 << k))
						in[i] += map[j][k];
	memset(f, -0x3f, sizeof(f));
	f[1][0] = 0;
	for (int i = 1; i < (1 << n); i++)
		for (int j = 0; j < n; j++)
			if (i & (1 << j))
			{
				int x = i ^ (1 << j);
				for (int k = 0; k < n; k++)
					if (i & (1 << k) && map[k][j])
						f[i][j] = max(f[i][j], f[x][k] + map[k][j]);
				for (int k = x; k; k = (k - 1) & x)
					f[i][j] = max(f[i][j], f[i ^ k][j] + in[1 << j ^ k]);
			}
	printf("%lld\n", in[(1 << n) - 1] - f[(1 << n) - 1][n - 1]);
}