Problem: 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 i and 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

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

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.

Solution (in JAVA):

Solution:


import sun.font.TrueTypeFont;

import java.util.Scanner;

/**
 * Created by t-hexiao on 2016/8/30.
 * https://hrbust-acm-team.gitbooks.io/acm-book/content/dynamic_programming/state.html
 */
public class Main {
    public static int Cnt(int st,int n) {
        int res = 0;
        for (int i = 0; i < n; i++) {
            if ((st&(1<<i)) != 0) {
                res++;
            }
        }
        return res;
    }
    public static long max(long a, long b){
        return a>b?a:b;
    }
    public static void main(String []args){
        Scanner in = new Scanner(System.in);
        int n  = in.nextInt();
        int m = in.nextInt();
        int k = in.nextInt();
        int[]a = new int[n];
        for(int i =0; i < n; i++){
            a[i] = in.nextInt();
        }
        int [][]c = new int[n][n];
        for(int i =0; i < k; i++){
            c[in.nextInt()-1][in.nextInt()-1] = in.nextInt();
        }
        int all = 1<<n;
        long ret = 0;
        long[][]dp = new long[all][n]; //如何表示压缩状态
        //如何表示状态的转移
        //遍历所有的状态？dp[i][st] 表示最后一个要选择 i时，目前状态能够获得的最大值
        for(int st =0; st < all; st++){ //注意遍历的方法，对所有的状态进行遍历
            for(int i =0; i < n; i++){
                if((st&(1<<i))!= 0){ //如果状态在第i位上为1，意味着这个碟子被拿了
                    boolean has = false;
                    for(int j =0; j < n; j++){

                        if(j!=i && (st&(1<<j)) != 0){ //j也被拿了
                            has = true;
                            //dp[st^(1<<i)][j]+a[i]+c[j][i] 表示 该状态下， 把第i置为0，然后在这种情况下，最后一个放置的如果是 j，那么如下
                            //相当于遍历了所有i之前是第j个状态的情况
                            dp[st][i] = max(dp[st][i],dp[st^(1<<i)][j]+a[i]+c[j][i]);
                        }
                    }
                    //
                    //如果此状态下没有其他碟子
                    if(!has){
                        dp[st][i] = a[i];
                    }
                }
                if(Cnt(st,n) == m){
                    ret = max(ret,dp[st][i]); //放放满盘子的状态
                }
            }
        }
        System.out.println(ret);
    }
}