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 n potions.
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 (15 potions 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.
题目描述
Anton玩一个非常interesting的游戏，可是他还差n个potions就可以到达人生巅峰了，如果不使用spells（奇淫巧技），那么Anton每x秒就会产出1个potions，现在给你两种Spells，第一种可以将生产1Potions的时间变为ai，消耗的能力值为bi；第二种将话费di能量直接产生ci的Potions，每种Speels至多使用一次，能量的上限值为s,现在问你Anton升到下一级所要花的最短时间。（施展奇淫巧技不需要时间）
思路
因为第二种Spells产生的Potions值和消耗的能量值是不减的，也就是说消耗的能力值越多威力越大，所以枚举第一种Spells，二分第二种Spells，最终的最小花费时间取最小值即可。
代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
typedef long long LL;
using namespace std;
const int maxn=1e9+7;
struct proc
{
LL v,c;
}a[200000+5],b[200000+5];
int cmp(proc a,proc b)
{
return a.c<b.c;
}
int main()
{
LL n,m,k,x,s;
while(~scanf("%lld %lld %lld",&n,&m,&k))
{
scanf("%lld %lld",&x,&s);
LL ans=n*x;
a[0].v=x;
a[0].c=0;
for(int i=1;i<=m;i++)
{
scanf("%lld",&a[i].v);
}
for(int i=1;i<=m;i++)
{
scanf("%lld",&a[i].c);
}
for(int i=1;i<=k;i++)
{
scanf("%lld",&b[i].v);
}
for(int i=1;i<=k;i++)
{
scanf("%lld",&b[i].c);
}
sort(a+1,a+1+m,cmp);
for(int i=0;i<=m;i++)
{
if(a[i].c>s) break;
int lest=s-a[i].c;
int l=1,r=k;
LL maxx=0;
while(l<=r)
{
int mid=(l+r)/2;
if(b[mid].c<=lest)
{
maxx=max(maxx,b[mid].v);
l=mid+1;
}
else r=mid-1;
}
if(n-maxx<=0) ans=0;
else ans=min(ans,(LL)(n-maxx)*a[i].v);
}
printf("%lld\n",ans);
}
return 0;
}