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 <bits/stdc++.h>
using namespace std;
const int N = 17; 
const int inf = 1e9; 
int n, m, a[N][N], f[1 << N][N], sum[1 << N]; 
int main()
{
   cin >> n >> m;
   for (int i = 1; i <= m; i++) {
      int u, v, z; 
      cin >> u >> v >> z;
      u--; v--; 
      a[u][v] = a[v][u] = z;
   }
   
   for (int i = 1; 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))
		  sum[i] += a[j][k];

   for (int i = 1; i < (1 << n); i++)
      for (int j = 0; j < n; j++)
	 f[i][j] = -inf;
   
   for (int i = 1; i < (1 << n); i++)
      if (i & 1) f[i][0] = sum[i]; 

   for (int i = 1; i < (1 << n); i++)
      for (int j = 0; j < n; j++)
	 if (i & (1 << j))
	    for (int k = (i - 1) & i; k > 0; k = (k - 1) & i)
	       if ((k & (1 << j)) == 0)
		  for (int l = 0; l < n; l++)
		     if ((k & (1 << l)) > 0 && a[j][l] > 0)
			f[i][j] = max(f[i][j], f[k][l] + a[j][l] + sum[i ^ k]);

   cout << sum[(1 << n) - 1] - f[(1 << n) - 1][n - 1] << endl; 
   return 0; 
}