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 <bitset>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int maxn = 15;
const int inf = 0x3f3f3f3f;

int mmap[maxn][maxn];
LL in[1 << maxn];
LL dp[1 << maxn][maxn];
int n, m;

int main()
{
	scanf("%d%d", &n, &m);
	for (int i = 0, f, t, d; i < m; ++i)
	{
		scanf("%d%d%d", &f, &t, &d);
		mmap[t - 1][f - 1] = mmap[f - 1][t - 1] = d;
	}
	for (int i = 0; i < (1 << n); ++i)
		for (int j = 0; j < n; ++j)
			if (i & (1 << j))
				for (int k = j + 1; k < n; ++k)
					if (i & (1 << k))
						in[i] += mmap[j][k];
	for (int i = 0; i < (1 << n); ++i)
		for (int j = 0; j < n; ++j)
			dp[i][j] = -inf;
	for (int i = 0; i < (1 << n); ++i)
	{
		for (int j = 0; j < n; ++j)
		{
			if (i & (1 << j))
			{
				if (!j)
					dp[i][j] = in[i];
				else
				{
					for (int k = 0; k < n; ++k)
						if (j != k && (i & (1 << k)) && mmap[k][j])
							dp[i][j] = max(dp[i][j], dp[i ^ (1 << j)][k] + mmap[k][j]);
					for (int tmp = (i ^ (1 << j)), k = tmp; k; k = (k - 1) & tmp)
						dp[i][j] = max(dp[i][j], dp[i ^ k][j] + in[k | (1 << j)]);
				}
			}
		}
	}
//	 cout << in[(1 << n) - 1] << endl;
	printf("%lld\n", in[(1 << n) - 1] - dp[(1 << n) - 1][n - 1]);
	return 0;
}