CodeForces 734C - Anton and Making Potions（枚举）

C. Anton and Making Potions

time limit per test4 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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 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.

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.

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.

题意：

要造n瓶毒药，每瓶需要x时间。但是你有s点魔力值。m个一类魔法，k个二类魔法。

一类魔法能改变制造的时间，二类魔法能直接产生一些毒药。

但是每一类最多只能使用一个。

问最小的时间。

解题思路：

枚举第一类魔法，在第一种魔法的条件下找出不大于当前魔力值的，消耗最多的第二类魔法（因为第二类魔法的功效和消耗都是非递减的，因此用二分去做）。

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+5;
int a_time[maxn];
int a_cost[maxn];
int b_make[maxn];
int b_cost[maxn];
int main()
{
int n,m,k;
int x,s;
cin>>n>>m>>k>>x>>s;
for(int i = 1;i <= m;i++)   cin>>a_time[i];
for(int i = 1;i <= m;i++)   cin>>a_cost[i];
for(int i = 1;i <= k;i++)   cin>>b_make[i];
for(int i = 1;i <= k;i++)   cin>>b_cost[i];
a_time[0] = x;
a_cost[0] = 0;
long long ans = 1LL*n*x;
for(int i = 0;i <= m;i++)
{
if(a_cost[i] > s)   continue;
int index = upper_bound(b_cost,b_cost+k+1,s-a_cost[i])-b_cost;
long long tmp = 1LL*(n-b_make[index-1])*a_time[i];
ans = min(ans,tmp);
}
cout<<ans<<endl;
return 0;
}