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>
using namespace std;

const int N = 16;

int dp[N][1 << N], sum[1 << N], G[N][N];

int main() {
	int n, m, res = 0; scanf("%d%d", &n, &m);
	for (int i = 1, a, b, c; i <= m; i++)
		scanf("%d%d%d", &a, &b, &c), G[a][b] = G[b][a] = c, res += c;
	for (int i = 1; i < 1 << n; i++)
		for (int j = 1; j <= n; j++) if (i & (1 << j - 1))
			for (int k = j + 1; k <= n; k++) if (i & (1 << k - 1))
				sum[i] += G[j][k];
	memset(dp, -0x3f, sizeof(dp));
	for (int s = 1; s < 1 << n; s++) if (s & 1) dp[1][s] = sum[s];
	for (int s = 1; s < 1 << n; s++)
		for (int i = 1; i < n; i++) if (dp[i][s] >= 0) {
			for (int j = 1; j <= n; j++) if (G[i][j] && !(s & (1 << j - 1))) {
				int t = (1 << n) - 1; t ^= s, t ^= (1 << j - 1);
				for (int t1 = t; ~t1; t1 = (t1 - 1) & t) {
					dp[j][s | (1 << j - 1) | t1] = max(dp[j][s | (1 << j - 1) | t1], 
						dp[i][s] + G[i][j] + sum[t1 | (1 << j - 1)]);
					if (!t1) break;
				}	
			}	
		}
	printf("%d\n", res - dp[n][(1 << n) - 1]);
}