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>
#define mov(x) (1<<(x))

using namespace std;

template <typename T> void cmin(T &x, const T &y)
{
	if(y < x) x = y;
}

template <typename T> void cmax(T &x, const T &y)
{
	if(y > x) x = y;
}

template <typename T> void read(T &x)
{
	x = 0; char c = getchar(); bool f = 0;
	while(!isdigit(c) && c!='-') c = getchar();
	if(c == '-') f = 1, c = getchar();
	while(isdigit(c)) x = x*10+c-'0', c = getchar();
	if(f) x = -x;
}

int n, m, sum;
int grp[mov(15)];
int w[15][15];
int f[mov(15)][15];

int main()
{
	read(n), read(m);
	memset(w, 0x9f, sizeof(w));
	for(int i=1; i<=m; i++)
	{
		int a, b, c;
		read(a), read(b), read(c);
		w[a-1][b-1] = c;
		w[b-1][a-1] = c;
		sum += c;
	}
	memset(f, 0x9f, sizeof(f));
	for(int s=1; s<mov(n); s++)
		for(int i=0; i<n; i++)
			if(s & mov(i))
				for(int j=i+1; j<n; j++)
					if(s & mov(j))
						if(w[i][j] > 0)
							grp[s] += w[i][j];
	for(int s=1; s<mov(n); s++)
		if(s & 1)
			f[s][0] = grp[s];
	for(int s=1; s<mov(n); s++)
		for(int i=0; i<n; i++)
			if(mov(i) & s)
			{
				int u = ((~s)&(mov(n)-1));
				for(int t=u; t; t=((t-1)&u))
					for(int j=0; j<n; j++)
						if(mov(j) & t)
							if(w[i][j] > 0)
								cmax(f[s|t][j], f[s][i] + w[i][j] + grp[t]);
			}
	printf("%d\n", sum-f[mov(n)-1][n-1]);
	return 0;
}