D. Kefa and Dishes

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ai 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 ai, (0 ≤ ai ≤ 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 xi, yi and ci (1 ≤ xi, yi ≤ n, 0 ≤ ci ≤ 109). That means that if you eat dish xi right before dish yi, then the Kefa's satisfaction increases by ci. It is guaranteed that there are no such pairs of indexes iand j (1 ≤ i < j ≤ k), that xi = xj and yi = yj.

Output

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

Examples

input

Copy

2 2 1
1 1
2 1 1

output

Copy

3

input

Copy

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

output

Copy

12

Note

In the first sample it is best to first eat the second dish, then the first one. Then we get one unit of satisfaction for each dish and plus one more for the rule.

In the second test the fitting sequences of choice are 4 2 1 or 2 1 4. In both cases we get satisfaction 7 for dishes and also, if we fulfill rule 1, we get an additional satisfaction 5.

題目大意：在n個菜裡選m個吃，每個菜有一個權值，另外有k個加成，形式為i j w，表示如果先吃了i，下一個吃的是j，那麼有權值w的加成，求最大權值

狀壓dp dp[state][i]表示狀態為state，最後吃的一個菜為i的最大權值。

#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;

#define ll long long 

const int N = 19;
ll dp[1<<N][N];
int val[N], mp[N][N];
int n, m, k;

int n_dish(int state) {
    int num = 0;
    for (int i = 0; i < n; i++) {
        if ((1<<i)&state) num++;
    }
    if (num == m) return 1;
    else return 0;
}

int main() {
    int x, y, c;
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 0; i < n; i++) scanf("%d", &val[i]);
    for (int i = 0; i < k; i++) {
        scanf("%d%d%d", &x, &y, &c);
        x--;y--;
        mp[x][y] = c;
    }
    ll ans = 0;
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < n; i++) dp[1<<i][i] = val[i];
    for (int i = 0; i < (1 << n); i++) {
        for (int j = 0; j < n; j++) {
            if ((1 << j)&i) {
                for (int k = 0; k < n; k++) {
                    if (k != j && (1 << k)&i) {
                        int state = i - (1 << j);
                        dp[i][j] = max(dp[i][j], dp[state][k] + mp[k][j] + val[j]);
                    }
                }
            }
        }
        if (n_dish(i)) {
            for (int ii = 0; ii < n; ii++) {
                ans = max(ans, dp[i][ii]);
            }
        }
    }
    printf("%I64d\n",ans);
    return 0;
}