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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

template <class T>
inline void relax(T &a, const T &b)
{
	if (b > a)
		a = b;
}

const int MaxN = 15;
const int INF = 0x3f3f3f3f;

int n, m;
int mat[MaxN][MaxN];

int sum[1 << MaxN];

int result = -INF;

int f[1 << MaxN][MaxN];

int main()
{
	cin >> n >> m;
	for (int i = 0; i < m; ++i)
	{
		int u, v, w;
		cin >> u >> v >> w, --u, --v;

		mat[u][v] = w;
		mat[v][u] = w;
	}

	int all = 1 << n;
	for (int s = 1; s < all; ++s)
	{
		int u = __builtin_ctz(s);

		sum[s] = sum[s ^ (s & -s)];
		for (int v = u + 1; v < n; ++v)
			if (s >> v & 1)
				sum[s] += mat[u][v];
	}

	for (int s = 0; s < all; ++s)
		for (int u = 0; u < n; ++u)
			f[s][u] = -INF;
	f[1][0] = 0;

	for (int s = 1; s < all; ++s)
	{
		for (int u = 0; u < n; ++u)
		{
			if (f[s][u] < 0)
				continue;

			int ex = (all - 1) ^ s;
			for (int t = ex; t; --t &= ex)
				relax(f[s | t][u], f[s][u] + sum[t | 1 << u]);

			for (int v = 0; v < n; ++v)
				if (mat[u][v] && ~s >> v & 1)
					relax(f[s | 1 << v][v], f[s][u] + mat[u][v]);
		}

		relax(result, f[s][n - 1]);
	}

	cout << sum[all - 1] - result << endl;

	return 0;
}