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.

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.

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

  1 #include<map>

  2

  3 #include<set>

  4

  5 #include<stack>

  6

  7 #include<cmath>

  8

  9 #include<queue>

 10

 11 #include<bitset>

 12

 13 #include<math.h>

 14

 15 #include<vector>

 16

 17 #include<string>

 18

 19 #include<stdio.h>

 20

 21 #include<cstring>

 22

 23 #include<iostream>

 24

 25 #include<algorithm>

 26

 27 #pragma comment(linker, "/STACK:102400000,102400000")

 28

 29 using namespace std;

 30

 31 typedef double db;

 32

 33 typedef long long ll;

 34

 35 typedef unsigned int uint;

 36

 37 typedef unsigned long long ull;

 38

 39 const db eps=1e-5;

 40

 41 const int N=2e5+10;

 42

 43 const int M=4e6+10;

 44

 45 const ll MOD=1000000007;

 46

 47 const int mod=1000000007;

 48

 49 const int MAX=1000000010;

 50

 51 const double pi=acos(-1.0);

 52

 53 ll a[N],b[N],c[N],d[N];

 54

 55 int main()

 56

 57 {

 58

 59     int i,m,k,l,r,mid;

 60

 61     ll n,x,s,t,ans;

 62

 63     scanf("%I64d%d%d", &n, &m, &k);

 64

 65     scanf("%I64d%I64d", &x, &s);

 66

 67     for (i=1;i<=m;i++) scanf("%I64d", &a[i]);

 68

 69     for (i=1;i<=m;i++) scanf("%I64d", &b[i]);

 70

 71     for (i=1;i<=k;i++) scanf("%I64d", &c[i]);

 72

 73     for (i=1;i<=k;i++) scanf("%I64d", &d[i]);

 74

 75

 76     ans=n*x;

 77

 78     for (i=1;i<=k;i++)

 79

 80     if (d[i]<=s) ans=min(ans,x*max(0ll,n-c[i]));

 81

 82

 83     for (i=1;i<=m;i++)

 84

 85     if (b[i]<=s&&a[i]<x) {

 86

 87         t=s-b[i];

 88

 89         if (t<d[1]) ans=min(ans,n*a[i]);

 90

 91         else {

 92

 93             l=1;r=k+1;mid=(l+r)>>1;

 94

 95             while (l+1<r)

 96

 97             if (d[mid]<=t) l=mid,mid=(l+r)>>1;

 98

 99             else r=mid,mid=(l+r)>>1;

100

101             ans=min(ans,max(0ll,n-c[l])*a[i]);

102

103         }

104

105     }

106

107     printf("%I64d\n", ans);

108

109     return 0;

110

111 }