#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T>
inline void printsp(T first) {
  print(first);
  putchar(' ');
}
template <class T>
inline void println(T first) {
  print(first);
  putchar('\n');
}
template <class T, class U>
inline void print(T first, U second) {
  printsp(first);
  println(second);
}
template <class T, class U, class V>
inline void print(T first, U second, V z) {
  printsp(first);
  printsp(second);
  println(z);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
int sum[1010], c[1010], cost[1010];
long long dp[1010][10100];
int main() {
  int n, W, B, X;
  gn(n, W);
  gn(B, X);
  for (int i = 1; i <= n; i++) gn(c[i]), sum[i] = sum[i - 1] + c[i];
  for (int i = 1; i <= n; i++) gn(cost[i]);
  memset(dp, -1, sizeof dp);
  dp[0][0] = W;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= sum[i]; j++)
      if (dp[i][j] >= 0) {
        long long val = min(W + (long long)B * j, dp[i][j] + X);
        for (int k = 0; k <= c[i + 1]; k++) {
          if (val < (long long)cost[i + 1] * k) break;
          smax(dp[i + 1][j + k], val - (long long)cost[i + 1] * k);
        }
      }
  }
  for (int i = sum[n]; i >= 0; i--)
    if (dp[n][i] >= 0) return println(i), 0;
  return 0;
}
