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 <cstring>
#include <algorithm>

using namespace std;
const int Max_N(15);

inline void update(int &a, const int &b)
{
	a = max(a, b);
}

int N, G[Max_N][Max_N], Sum[1 << Max_N];

void init()
{
	int M;
	scanf("%d%d", &N, &M);
	int a, b, c;
	while (M--)
		scanf("%d%d%d", &a, &b, &c), --a, --b, G[a][b] = G[b][a] = c;
	for (int S = 0;S < (1 << N);++S)
		for (int i = 0;i <= N - 1;++i)
			for (int j = i + 1;j <= N - 1;++j)
				if ((S & (1 << i)) && (S & (1 << j)))
					Sum[S] += G[i][j];
}

//等价于最大化保留的边权和，使得1 ~ n的路径上所有边都是桥边
//最优方案一定是一个1 ~ n的路径上，每个点挂一些集合 
//F[S][i]表示目前dp到1 ~ i的路径，满足1 ~ i的路径上所有边都是桥边，用到的点的集合是S
//转移考虑增加一条桥边还是新挂一个集合
int F[1 << Max_N][Max_N]; 
void dp()
{
	memset(F, -0X3F, sizeof(F));
	F[1 << 0][0] = 0;
	for (int S = 0;S < (1 << N);++S)
		for (int last = 0;last <= N - 1;++last)
			if (S & (1 << last))
			{
				for (int now = 0;now <= N - 1;++now)
					if ((S & (1 << now)) == 0 && G[last][now])
						update(F[S | (1 << now)][now], F[S][last] + G[last][now]);
				for (int S0 = ((1 << N) - 1) - S;S0;S0 = ((S0 - 1) & ((1 << N) - 1 - S)))
					update(F[S | S0][last], F[S][last] + Sum[S0 + (1 << last)]);
			}
	printf("%d", Sum[(1 << N) - 1] - F[(1 << N) - 1][N - 1]);
}

int main()
{
	init();
	dp();
	return 0;
}