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 <algorithm>
int N, M, g[15][15], f[32768][15];
int main()
{
	scanf("%d%d", &N, &M);
	for (int i = 0, u, v, w; i < M; i++)
	{
		scanf("%d%d%d", &u, &v, &w);
		u--, v--;
		g[u][v] = g[v][u] = w;
	}
	for (int i = 0; i < 1 << N; i++)
		for (int j = 0; j < N; j++)
			f[i][j] = 2000000000;
	for (int i = 1; i < 1 << N; i += 2)
		f[i][0] = 0;
	for (int i = 1; i < 1 << N; i++)
		for (int j = 0; j < N; j++)
			if (f[i][j] < 2000000000)
			{
				static int s[15];
				for (int k = 0; k < N; k++)
				{
					s[k] = 0;
					for (int l = 0; l < N; l++)
						if (i >> l & 1)
							s[k] += g[k][l];
				}
				for (int k = i; k < 1 << N; k = k + 1 | i)
				{
					int S = 0;
					for (int l = 0; l < N; l++)
						if (k - i >> l & 1)
							S += s[l];
					for (int l = 0; l < N; l++)
						if (k - i >> l & 1 && g[j][l])
							f[k][l] = std::min(f[k][l], f[i][j] + S - g[j][l]);
				}
			}
	printf("%d\n", f[(1 << N) - 1][N - 1]);
	return 0;
}