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 INF = 1e9;
const int maxn = 16;
const int maxs = 1 << maxn;
int a[maxn][maxn];
int f[maxs], g[maxn][maxs];
int tot, n, m, S;
void calc(int i, int s)
{
	for(int j = 0; j < n; ++j)
		if((s & (1 << j)) && a[i][j])
			g[i][s] = max(g[i][s], g[j][s - (1 << i)] + a[i][j]);
	int res = s - (1 << i);
	for(int r = res; r; r = res & (r - 1))
		g[i][s] = max(g[i][s], g[i][r | (1 << i)] + f[s - r]);
	g[i][s] = max(g[i][s], g[i][1 << i] + f[s]);
	//cout << i << ' ' << s << ' ' << g[i][s] << endl;
}
int main()
{
	scanf("%d%d", &n, &m); 
	S = (1 << n) - 1;
	for(int i = 1; i <= m; ++i)
	{
		int u, v, w;
		scanf("%d%d%d", &u, &v, &w);
		--u; --v;
		a[u][v] = a[v][u] = w;
		tot += w;
	}
	for(int j = 0; j < n; ++j)
	{
		int cur = 1 << j;
		for(int s = 0; s < cur; ++s)
		{
			f[s | cur] = f[s];
			for(int i = 0; i < j; ++i)
				if(s & (1 << i))
					f[s | cur] += a[i][j];
		}
	}
	for(int s = 1; s <= S; ++s)
		for(int i = 0; i < n; ++i)
			g[i][s] = -INF;
	g[0][1] = 0;
	for(int s = 1; s <= S; ++s)
		for(int i = 0; i < n; ++i)
			if(s & (1 << i))
				calc(i, s);
	//cout << g[n - 1][S] << endl;
	printf("%d\n", tot - g[n - 1][S]);
	return 0;
}
/*
4 4
1 2 100
2 4 100
1 3 100
3 4 100
*/