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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
int dp[1 << 15][15];
int cnt[1 << 15];
int m[15][15];
int main()
{
	int num, way;
	scanf("%d%d", &num, &way);
	int sum = 0;
	for (int i = 0; i < way; i++)
	{
		int za, zb, zc;
		scanf("%d%d%d", &za, &zb, &zc);
		za--, zb--;
		m[za][zb] = m[zb][za] = zc;
		sum += zc;
	}
	for (int p = 0; p < (1 << num); p++)
	{
		for (int i = 0; i < num; i++)
		{
			for (int j = i + 1; j < num; j++)
			{
				if ((p&(1 << i)) != 0 && (p&(1 << j)) != 0)cnt[p] += m[i][j];
			}
		}
	}
	for (int i = 0; i < (1 << num); i++)for (int j = 0; j < num; j++)dp[i][j] = -1000000000;
	for (int p = 0; p < (1 << num); p++)if (p & 1)dp[p][0] = cnt[p];
	for (int p = 0; p < (1 << num); p++)
	{
		for (int i = 0; i < num; i++)
		{
			if ((p&(1 << i)) == 0)continue;
			for (int j = 0; j < num; j++)
			{
				if (p&(1 << j))continue;
				if (m[i][j] == 0)continue;
				int mask = (1 << num) - 1 - p - (1 << j);
				int now = mask;
				for (;;)
				{
					dp[p | now | (1 << j)][j] = max(dp[p | now | (1 << j)][j], dp[p][i] + m[i][j] + cnt[now | (1 << j)]);
					if (now == 0)break;
					now--;
					now &= mask;
				}
			}
		}
	}
	printf("%d\n", sum - dp[(1 << num) - 1][num - 1]);
}