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 <cstring>
#include <cstdio>
int val[15][15], sum[100005], dp[100005][15]; 
inline void upd(int &x, int y)
{
	if (x < y)
		x = y; 
}
int main()
{
	// freopen("ARC078-F.in", "r", stdin); 
	int n, m; 
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; i++)
	{
		int u, v, w; 
		scanf("%d%d%d", &u, &v, &w); 
		val[--u][--v] = w; 
		val[v][u] = w; 
	}
	for (int i = 0; i < 1 << n; i++)
	{
		for (int x = 0; x < n; x++)
		{
			for (int y = x; y < n; y++)
			{
				if ((i & (1 << x)) && (i & (1 << y)))
					sum[i] += val[x][y]; 
			}
		}
	}
	memset(dp, -0x3f, sizeof(dp)); 
	dp[1][0] = 0; 
	for (int s = 0; s < 1 << n; s++)
	{
		for (int i = 0; i < n; i++)
		{
			if (!(s & (1 << i)))
				continue; 
			for (int j = 0; j < n; j++)
			{
				if ((s & (1 << j)) || !val[i][j])
					continue; 
				upd(dp[s | (1 << j)][j], dp[s][i] + val[i][j]); 
			}
			int x = (1 << n) - 1 ^ s; 
			for (int t = x; t; t = t - 1 & x)
				upd(dp[s | t][i], dp[s][i] + sum[t | (1 << i)]); 
		}
	}
	printf("%d\n", sum[(1 << n) - 1] - dp[(1 << n) - 1][n - 1]);
	return 0; 
}