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 <cstdlib>
#include <algorithm>
#include <cstring>

const int maxn = 20;
const int maxs = (1 << 15) + 1;
const int inf = 0x3f3f3f3f;

int n, m;
int G[maxn][maxn];

int fs;
int f[maxs][maxn];
int g[maxs][maxn];

void checkmin(int &x,int y)
{
	if(y < x) x = y;
}
int main()
{
	scanf("%d%d", &n, &m);
	for(int i = 1, a, b, c; i <= m; i++)
	{
		scanf("%d%d%d", &a, &b, &c);
		--a, --b;
		G[a][b] = G[b][a] = c;
	}

	fs = (1 << n) - 1;
	
	for(int S = 1; S <= fs; S++)
		for(int i = 0; i < n; i++)
		{
			if((S >> i) & 1) continue;
			
			for(int j = 0; j < n; j++)
				if((S >> j) & 1) g[S][i] += G[i][j];
		}
	
	memset(f, inf, sizeof(f));
	
	f[1][0] = 0;
	for(int S = 1; S <= fs; S++)
		for(int i = 0; i < n; i++)
			if((S >> i) & 1)
			{
				for(int s = (S - 1) & S; s; s = (s - 1) & S)
					if((s >> i) & 1)
					{
						int val = 0;
					
						for(int j = 0; j < n; j++)
							if(i != j && ((s >> j) & 1))
								val += g[S ^ s][j];
						
						checkmin(f[S][i], f[s][i] + val);
					}
				
				for(int j = 0; j < n; j++)
					if((S >> j) & 1)
					{
						if(!G[j][i]) continue;
						
						int s = S ^ (1 << i);
						
						checkmin(f[S][i], f[s][j] + g[s ^ (1 << j)][i]);
					}
			}
	
	printf("%d\n", f[fs][n - 1]);
	return 0;
}