题目链接：https://cn.vjudge.net/problem/ZOJ-4019

DreamGrid has a magical knapsack with a size capacity of  called the Schrödinger's knapsack (or S-knapsack for short) and two types of magical items called the Schrödinger's items (or S-items for short). There are  S-items of the first type in total, and they all have a value factor of k1; While there are  S-items of the second type in total, and they all have a value factor of k2. The size of an S-item is given and is certain. For the -th S-item of the first type, we denote its size by  ; For the -th S-item of the second type, we denote its size by ;.

But the value of an S-item remains uncertain until it is put into the S-knapsack (just like Schrödinger's cat whose state is uncertain until one opens the box). Its value is calculated by two factors: its value factor k, and the remaining size capacity r of the S-knapsack just after it is put into the S-knapsack. Knowing these two factors, the value  of an S-item can be calculated by the formula v=kr.

For a normal knapsack problem, the order to put items into the knapsack does not matter, but this is not true for our Schrödinger's knapsack problem. Consider an S-knapsack with a size capacity of 5, an S-item with a value factor of 1 and a size of 2, and another S-item with a value factor of 2 and a size of 1. If we put the first S-item into the S-knapsack first and then put the second S-item, the total value of the S-items in the S-knapsack is 1*（5-2）+2*（3-1）=7; But if we put the second S-item into the S-knapsack first, the total value will be changed to 2*（5-1）+1*（4-2）=10. The order does matter in this case!

Given the size of DreamGrid's S-knapsack, the value factor of two types of S-items and the size of each S-item, please help DreamGrid determine a proper subset of S-items and a proper order to put these S-items into the S-knapsack, so that the total value of the S-items in the S-knapsack is maximized.

Input

The first line of the input contains an integer T (about 500), indicating the number of test cases. For each test case:

The first line contains three integers k1,k2  and  c(1<=k1,k2,c<=1e7), indicating the value factor of the first type of S-items, the value factor of the second type of S-items, and the size capacity of the S-knapsack.

The second line contains two integers  n and m (1<=n,m<=2000), indicating the number of the first type of S-items, and the number of the second type of S-items.

The next line contains n integers  , indicating the size of the S-items of the first type.

The next line contains m integers  , indicating the size of the S-items of the second type.

It's guaranteed that there are at most 10 test cases with their max(n,m) larger than 100.

Output

For each test case output one line containing one integer, indicating the maximum possible total value of the S-items in the S-knapsack.

Sample Input

3
3 2 7
2 3
4 3
1 3 2
1 2 10
3 4
2 1 2
3 2 3 1
1 2 5
1 1
2
1
Sample Output

23
45
10
Hint

For the first sample test case, you can first choose the 1st S-item of the second type, then choose the 3rd S-item of the second type, and finally choose the 2nd S-item of the first type. The total value is 2*（7-1）+2*（6-2）+3*（4-3）=23.

For the second sample test case, you can first choose the 4th S-item of the second type, then choose the 2nd S-item of the first type, then choose the 2nd S-item of the second type, then choose the 1st S-item of the second type, and finally choose the 1st S-item of the first type. The total value is 2*（10-1）+1*（9-1）+2*（8-2）+2*（6-3）+1*（3-2）=45.

The third sample test case is explained in the description.

It's easy to prove that no larger total value can be achieved for the sample test cases.

思路：考虑只有一种类型物品的时候，肯定是先选体积小的物品最优。考虑2种类型，则需要用d[i][j]表示第一种类型的物品选体积前i小的，第二类物品选体积前j小时的最大价值。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=2e3+5;
ll a[N],b[N],dp[N][N],sum1[N],sum2[N];
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		ll k1,k2,c;
		scanf("%lld%lld%lld",&k1,&k2,&c);
		int n,m;
		scanf("%d%d",&n,&m);
		for(int i=1;i<=n;i++)
		scanf("%lld",&a[i]);
		sort(a+1,a+1+n);
		for(int i=1;i<=n;i++)
		sum1[i]=sum1[i-1]+a[i];
		for(int i=1;i<=m;i++)
		scanf("%lld",&b[i]);
		sort(b+1,b+1+m);
		for(int i=1;i<=m;i++)
		sum2[i]=sum2[i-1]+b[i];
		ll ans=0;
		for(int i=0;i<=n;i++){
			for(int j=0;j<=m;j++){
				dp[i][j]=0;
				if(sum1[i]+sum2[j]>c) continue;
				if(i) dp[i][j]=max(dp[i][j],dp[i-1][j]+k1*(c-sum1[i]-sum2[j]));
				if(j) dp[i][j]=max(dp[i][j],dp[i][j-1]+k2*(c-sum1[i]-sum2[j]));
				ans=max(ans,dp[i][j]);
			}
		}
		printf("%lld\n",ans);
	}
	return 0;
}