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 java.util.LinkedList;
import java.util.Scanner;

/**
 * 状态压缩dp
 * 少于1000次输入，用Scanner
 */
public class Main {
	static int n,m,k;
	static int[] a;
	static int[][] val;
	static long[][] dp;
	static long max;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		m = sc.nextInt();
		k = sc.nextInt();
		a = new int[20];
		for(int i = 0; i < n; i++)
			a[i] = sc.nextInt();
		dp = new long[1000000][20];//最大有2^18种状态
		val = new int[20][20];
		for(int i = 0; i < k; i++){//全部从0-n-1计数
			int x = sc.nextInt();
			int y = sc.nextInt();
			int c = sc.nextInt();
			val[x-1][y-1] = c;
		}
		sc.close();
		int end = 1<<n;
		for(int st = 0; st < end; st++){
			//对每一个状态st，找其之前的状态来做dp
			if(cnt(st)>m)
				continue;
			for(int i = 0; i < n; i++){
				if((st&(1<<i))>0){//若点了i
					boolean has = false;
					for(int j = 0; j < n; j++){
						if(i!=j && (st&(1<<j))>0){//如果还点了j
							dp[st][i] = Math.max(dp[st][i], 
									dp[st&(~(1<<i))][j]+val[j][i]+a[i]);
							//和没有点i但是点了j的上一单进行dp推进
							has = true;
						}
					}
					if(!has){//只有点一单i，其他都没点
						dp[st][i] = a[i];
					}
				}
				max = Math.max(max, dp[st][i]);
			}
		}
		System.out.println(max);
	}
/*	
	private static void dfs(int st, int end, int dep) {
		// TODO Auto-generated method stub
		//dfs全排列
		if(dep==0)
			return;
		for(int i = 0; i < n; i++){
			if((st&(1<<i))==0){//表示没有选择第i个菜
				dp[st|(1<<i)][i] = Math.max(dp[st|(1<<i)][i],
						dp[st][end]+val[end][i]+a[i]);
				//这里的dp[st][end]有可能不是最大值，所以错误！
				//不应该用dfs
				max = Math.max(max, dp[st|(1<<i)][i]);
				dfs(st|(1<<i), i, dep-1);
			}
		}
	}
*/
	private static int cnt(int st) {
		// TODO Auto-generated method stub
		int res = 0;
		while(st>0){
			res+=st&1;
			st>>=1;
		}
		return res;
	}
}