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;

const int MAXN = 20, INF = 1e9;

int n, m, x, y, c;

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

int cost(int x, int y) {
	return sum[x | y] - (sum[x] + sum[y]);
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; ++i) {
		scanf("%d%d%d", &x, &y, &c); x--; y--;
		G[x][y] = c;
		G[y][x] = c;
	}

	for (int i = 0; i < 1 << n; ++i) {
		for (int j = 0; j < n; ++j) {
			if (((i >> j)&1) == 0) continue;
			for (int k = j+1; k < n; ++k) {
				if ((i >> k)&1)	sum[i] += G[j][k];
			}
		}
	}

	for (int i = 0; i < 1 << n; ++i) {
		for (int j = 0; j < n; ++j) {
			dp[i][j] = INF;
		}
	}
	dp[1][0] = 0;

	for (int i = 0; i < 1 << n; ++i) {
		for (int j = 0; j < n; ++j) {
			if (dp[i][j] == INF) continue;
			for (int k = 0; k < n; ++k) {
				if ((i >> k) & 1) continue;
				if (G[j][k] == 0) continue;
				dp[i | (1 << k)][k] = min(dp[i | (1 << k)][k], 
						dp[i][j] + cost(i ^ (1 << j), 1 << k));
			}
			for (int k = ((1 << n)-1) & ~i; k > 0; k = (k-1)& ~i) {
				dp[i | k][j] = min(dp[i | k][j], dp[i][j] + cost(i ^ (1 << j), k));
			}
		}
	}

	printf("%d\n", dp[(1 << n)-1][n-1]);


	return 0;
}