580D

When Kefa came to the restaurant and sat at a table, the waiter immediately brought him the menu. There weren dishes. Kefa knows that he needs exactlym 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 hima_i units of satisfaction. But some dishes do not go well together and some dishes go very well together. Kefa set to himselfk rules of eating food of the following type — if he eats dishx exactly before dish y (there should be no other dishes between x and y), then his satisfaction level raises byc.

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 numbersa_i, (0 ≤ a_i ≤ 10⁹) — the satisfaction he gets from thei-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 dishx_i right before dishy_i, then the Kefa's satisfaction increases byc_i. It is guaranteed that there are no such pairs of indexesi and j (1 ≤ i < j ≤ k), thatx_i = x_j andy_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 11 12 1 1

Output

3

Input

4 3 21 2 3 42 1 53 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<stdio.h>#include<algorithm>#include<string.h>using namespace std;const int maxm = 1 << 20;#define ll long longll 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;}