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<algorithm>
using namespace std;
int D[1 << 15][15], S[1 << 15], U[1<<15][15];
int w[15][15], n, m, sum;
int main() {
	int i, a, b, c, j, k, l;
	scanf("%d%d", &n, &m);
	for (i = 0; i < m; i++) {
		scanf("%d%d%d", &a, &b, &c);
		a--, b--;
		w[a][b] = w[b][a] = c;
		sum += c;
		for (j = 0; j < (1 << n); j++) {
			if (((~j)&((1 << a) | (1 << b))) == 0)S[j] += c;
			if (j&(1 << a)) U[j][b] += c;
			if (j&(1 << b)) U[j][a] += c;
		}
	}
	for (i = 0; i < (1 << n); i++) {
		for (j = 0; j < n; j++)D[i][j] = -1e9;
	}
	D[1][0] = 0;
	for (i = 0; i < (1 << n); i++) {
		for (j = 0; j < n; j++) {
			if (D[i][j] < 0)continue;
			int b = (1 << n) - i - 1;
			for (k = b;k; k = (k - 1)&b) {
				D[i | k][j] = max(D[i | k][j], D[i][j] + S[k] + U[k][j]);
			}
			for (k = 0; k < n; k++) {
				if (i&(1 << k))continue;
				if (!w[j][k])continue;
				D[i | (1 << k)][k] = max(D[i | (1 << k)][k], D[i][j] + w[j][k]);
			}
		}
	}
	printf("%d\n", sum - D[(1 << n) - 1][n - 1]);
}