580D - Kefa and Dishes

When Kefa came to the restaurant and sat at a table, the waiter immediately brought him the menu. There were n dishes. Kefa knows that he needs exactly m dishes. But at that, he doesn't want to order the same dish twice to taste as many dishes as possible.

Kefa knows that the i-th dish gives him a_i units of satisfaction. But some dishes do not go well together and some dishes go very well together. Kefa set to himself k rules of eating food of the following type — if he eats dish x exactly before dish y (there should be no other dishes between x and y), then his satisfaction level raises by c.

Of course, our parrot wants to get some maximal possible satisfaction from going to the restaurant. Help him in this hard task!

Input

The first line of the input contains three space-separated numbers, n, m and k (1 ≤ m ≤ n ≤ 18, 0 ≤ k ≤ n * (n - 1)) — the number of dishes on the menu, the number of portions Kefa needs to eat to get full and the number of eating rules.

The second line contains n space-separated numbers a_i, (0 ≤ a_i ≤ 10⁹) — the satisfaction he gets from the i-th dish.

Next k lines contain the rules. The i-th rule is described by the three numbers x_i, y_i and c_i (1 ≤ x_i, y_i ≤ n, 0 ≤ c_i ≤ 10⁹). That means that if you eat dish x_i right before dish y_i, then the Kefa's satisfaction increases by c_i. It is guaranteed that there are no such pairs of indexes i and j (1 ≤ i < j ≤ k), that x_i = x_j and y_i = y_j.

Output

In the single line of the output print the maximum satisfaction that Kefa can get from going to the restaurant.

Examples

Input

2 2 1
1 1
2 1 1

Output

3

Input

4 3 2
1 2 3 4
2 1 5
3 4 2

Output

12

有n中菜肴，每种菜肴都有一个愉悦度，同时制定k个规则，当吃了第i中菜的下一道菜吃第j种菜则愉悦度增加g[i][j]。问你吃m种菜可获得的最大愉悦度。

dp[i][j]表示当前状态为i且最后吃第j种菜的最大愉悦度，则递推公式为dp[i | (1 << k)][k] = max(dp[i | (1 << k)][k], dp[i][j] + a[k] + g[j][k])。

#include
#include
#include
using namespace std;
const int maxm = 1 << 20;
#define ll long long
ll dp[maxm][25], g[25][25], a[maxm];
int main()
{
	int n, i, j, k, m, x, y, z, cnt, flag;
	ll ans;
	scanf("%d%d%d", &n, &m, &k);
	memset(dp, -1, sizeof(dp));
	for (i = 0;i < n;i++)
		scanf("%lld", &a[i]);
	for (i = 1;i <= k;i++)
	{
		scanf("%d%d%d", &x, &y, &z);
		x--, y--;
		g[x][y] = z;
	}
	for (i = 0;i < n;i++)
		dp[1 << i][i] = a[i];
	ans = 0;
	for (i = 0;i < (1 << n);i++)
	{
		ll now = i;
		cnt = 0;flag = 0;
		while (now)
		{
			if (now & 1) cnt++;
			now >>= 1;
		}
		if (cnt == m) flag = 1;
		for (j = 0;j < n;j++)
		{
			if (dp[i][j] == -1) continue;
			if (flag) ans = max(ans, dp[i][j]);
			for (k = 0;k < n;k++)
			{
				if (i&(1 << k)) continue;
				dp[i | (1 << k)][k] = max(dp[i | (1 << k)][k], dp[i][j] + a[k] + g[j][k]);
			}
		}
	}
	printf("%lld\n", ans);
	return 0;
}