CF580D_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

Input

Output

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

（别吐槽题面字大，我也没有找到更好的方法qwq，除非你们想看shi色的题面）

这题是类似背包的一种实现方式。
题目中的“两道菜的顺序先后组合的附加值”不能忽视，因为若无视顺序直接背包的话会出现类似于“环”的非法情况。所以为了记录顺序，考虑在本来背包一维的基础上再开一维，记录最近吃掉的菜的编号。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
#define Max(a,b) ((a>b)?a:b)
typedef long long LL;
const int N=(1<<19);
LL dp[N][20],ans=0;
int n,m,K,num[N],q[N],c[20][20],a[N],cnt=0;
inline int read(){
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9') {x=(x<<3)+(x<<1)+ch-48;ch=getchar();}
	return x*f;}
int main(){
	int i,j;
	n=read();m=read();K=read();
	memset(num,0,sizeof(num));memset(c,0,sizeof(c));
	memset(dp,0,sizeof(dp));
	for (i=1;i<=n;++i) a[i]=read();
	for (i=0;i<(1<<n);++i) {
		for (j=0;(1<<j)<=i;++j) if ((1<<j)&i) ++num[i];
		if (num[i]==m) q[++cnt]=i;}
	for (i=0;i<n;++i) dp[(1<<i)][i]=a[n-i];
	for (i=1;i<=K;++i) {
		int x=read(),y=read();c[n-x][n-y]=read();}
	for (i=0;i<(1<<n);++i){
		for (j=0;(1<<j)<=i;++j){
			if (i&(1<<j)){
				for (int k=0;k<n;++k){
					if (!(i&(1<<k))) dp[i|(1<<k)][k]=Max(dp[i|(1<<k)][k],dp[i][j]+c[j][k]+a[n-k]);
				}
			}
		}
	}
	for (i=1;i<=cnt;++i) for (j=0;j<n;++j){ans=Max(ans,dp[q[i]][j]);}
	printf("%lld
",ans);
	return 0;
}