Codeforces Round #379 (Div. 2) C. Anton and Making Potions//734C

C. Anton and Making Potions

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Anton is playing a very interesting computer game, but now he is stuck at one of the levels. To pass to the next level he has to prepare npotions.

Anton has a special kettle, that can prepare one potions in x seconds. Also, he knows spells of two types that can faster the process of preparing potions.

1. Spells of this type speed up the preparation time of one potion. There are m spells of this type, the i-th of them costs bi manapoints and changes the preparation time of each potion to ai instead of x.
2. Spells of this type immediately prepare some number of potions. There are k such spells, the i-th of them costs di manapoints and instantly create ci potions.

Anton can use no more than one spell of the first type and no more than one spell of the second type, and the total number of manapoints spent should not exceed s. Consider that all spells are used instantly and right before Anton starts to prepare potions.

Anton wants to get to the next level as fast as possible, so he is interested in the minimum number of time he needs to spent in order to prepare at least n potions.

Input

The first line of the input contains three integers n, m, k (1 ≤ n ≤ 2·109, 1 ≤ m, k ≤ 2·105) — the number of potions, Anton has to make, the number of spells of the first type and the number of spells of the second type.

The second line of the input contains two integers x and s (2 ≤ x ≤ 2·109, 1 ≤ s ≤ 2·109) — the initial number of seconds required to prepare one potion and the number of manapoints Anton can use.

The third line contains m integers ai (1 ≤ ai < x) — the number of seconds it will take to prepare one potion if the i-th spell of the first type is used.

The fourth line contains m integers bi (1 ≤ bi ≤ 2·109) — the number of manapoints to use the i-th spell of the first type.

There are k integers ci (1 ≤ ci ≤ n) in the fifth line — the number of potions that will be immediately created if the i-th spell of the second type is used. It's guaranteed that ci are not decreasing, i.e. ci ≤ cj if i < j.

The sixth line contains k integers di (1 ≤ di ≤ 2·109) — the number of manapoints required to use the i-th spell of the second type. It's guaranteed that di are not decreasing, i.e. di ≤ dj if i < j.

Output

Print one integer — the minimum time one has to spent in order to prepare n potions.

Examples

input

20 3 2
10 99
2 4 3
20 10 40
4 15
10 80

output

20

input

20 3 2
10 99
2 4 3
200 100 400
4 15
100 800

output

200

Note

In the first sample, the optimum answer is to use the second spell of the first type that costs 10 manapoints. Thus, the preparation time of each potion changes to 4 seconds. Also, Anton should use the second spell of the second type to instantly prepare 15 potions spending 80 manapoints. The total number of manapoints used is 10 + 80 = 90, and the preparation time is 4·5 = 20 seconds (15potions were prepared instantly, and the remaining 5 will take 4 seconds each).

In the second sample, Anton can't use any of the spells, so he just prepares 20 potions, spending 10 seconds on each of them and the answer is 20·10 = 200.

暴力枚举，加点优化，第二类型的药水给的数据都是严格递增的，所以切入点在这，枚举第二类型药水，枚举的过程中重点是找第一类型药水里花费符合条件的时间最小的那个药水（就是bi符合条件，ai最小），主要点应该就在找第一类型的药水，先按费用bi从小到大排序，在从前往后扫一遍，每个ai都存0~i里时间（a）最小的那个，因为 i 的费用（b）合适了，那么他前面费用比他小的  肯定也合适，所以存耗时最小，

。。数据范围是2*10^5，没看到2.。。数组开小了，交了也不报re。。报wa，一直调一直调，都没调对，第二天看了样例才知道。。。。很傻，很气

看别的博客二分非常简单，枚举第一类型，然后对第二类型二分

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node
{
	long long int a,b;
}num[200005];
long long int c[200005],d[200005];
long long int i,j,n,m,k,x,s,sum,minn=0,index,minindex;
int cmp(node u,node v)
{
		return u.b<v.b;
}
int main()
{
	int flog1=1;
	scanf("%lld %lld %lld",&n,&m,&k);
	scanf("%lld %lld",&x,&s);
	for(i=0;i<m;i++)
	scanf("%lld",&num[i].a);
	for(i=0;i<m;i++)
	scanf("%lld",&num[i].b);
	sort(num,num+m,cmp);
	for(i=1;i<=k;i++)
	scanf("%lld",&c[i]);
	for(i=1;i<=k;i++)
	scanf("%lld",&d[i]);
	d[0]=c[0]=0;
	index=m-1;
	for(i=1;i<m;i++)
	{
		if(num[i].a>num[i-1].a)
		num[i].a=num[i-1].a;
	}
	for(i=0;i<=k;i++)
	{	
		int flog=0;
		sum=0;
		long long int nn=n,xx=x,ss=s;
		if(i!=0&&c[i]==c[i-1]&&d[i]==d[i-1])
		continue;
		if(ss>=d[i])
		{
			nn-=c[i];
			ss-=d[i];	
		}
		else
		break;
		if(nn>0)
		{
			while(ss<num[index].b)
			index--;
			if(index>=0&&num[index].a<xx)
			{
				sum+=nn*num[index].a;
				nn=0;	
			}
			else
			{
				sum+=nn*x;
				nn=0;
			}
		}
		if(flog1)
		{
			minn=sum;
			flog1=0;
		}
		else if(sum<minn)
		minn=sum;			
	}
	printf("%lld\n",minn);
}