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 <iostream>
#include <algorithm>
using namespace std;
const int N = 15;
int mat[N][N], inside[1 << N], cross[1 << N][N], f[1 << N][N];
int main()
{
	int n, m;
	cin >> n >> m;
	while (m--)
	{
		int u, v, w;
		cin >> u >> v >> w;
		mat[u - 1][v - 1] = mat[v - 1][u - 1] = w;
	}
	for (int i = 0; i < 1 << n; i++)
		for (int j = 0; j < n; j++)
			if (i & (1 << j))
				for (int k = 0; k < j; k++)
					if (i & (1 << k))
						inside[i] += mat[j][k];
	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))
						cross[i][j] += mat[j][k];
	fill_n(&f[0][0], sizeof(f) / sizeof(int), -1);
	f[1][0] = 0;
	for (int i = 0; i < 1 << n; i++)
		for (int j = 0; j < n; j++)
			if (~f[i][j])
			{
				for (int k = 0; k < n; k++)
					if (!(i & (1 << k)) && mat[j][k])
						f[i | (1 << k)][k] = max(f[i | (1 << k)][k], f[i][j] + mat[j][k]);
				int c = (1 << n) - i - 1;
				for (int s = c; s; s = (s - 1) & c)
					f[i | s][j] = max(f[i | s][j], f[i][j] + inside[s] + cross[s][j]);
			}
	cout << inside[(1 << n) - 1] - f[(1 << n) - 1][n - 1] << endl;
	return 0;
}