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 = 16, inf = 2e9;

int n, m;
int f[N][1 << N], g[N][1 << N], a[1 << N];
int G[N][N], ans;

int main()
{
	scanf("%d %d", &n, &m);
	for(int i = 1; i <= m; i++)
	{
		int x, y, t;
		scanf("%d %d %d", &x, &y, &t);
		G[x][y] = G[y][x] = t;
		ans += t;
	}

	for(int s = 0; s < (1 << n); s++)
		for(int i = 1; i <= n; i++)
			if(s & (1 << i - 1))
				for(int j = i + 1; j <= n; j++)
					if(s & (1 << j - 1)) a[s] += G[i][j];

	for(int s = 0; s < (1 << n); s++)
		for(int i = 1; i <= n; i++) g[i][s] = f[i][s] = -inf;

	for(int s = 1; s < (1 << n); s++)
		if(s & 1) f[1][s] = a[s];
			
	for(int s = 1; s < (1 << n); s++)
	{
		if(!(s & 1)) continue;
		for(int i = 2; i <= n; i++)
			if(s & (1 << i - 1))
				for(int t = s; t; t = (t - 1) & s)
					if((t & (1 << i - 1)) && !(t & 1))
						f[i][s] = max(f[i][s], g[i][s ^ t] + a[t]);
		for(int i = 1; i <= n; i++)
			if(!(s & (1 << i - 1)))
				for(int j = 1; j <= n; j++)
					if(s & (1 << j - 1))
						if(G[i][j]) g[i][s] = max(g[i][s], f[j][s] + G[i][j]);
	}

	printf("%d", ans - f[n][(1 << n) - 1]);
	return 0;
}