Problem:
Score: 500 points

Problem Statement

Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.

In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:

A team should assign one member to each food, and should not assign the same member to multiple foods.

It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.

The score of a team is the longest time it takes for an individual member to finish the food.

Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.

What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?

Constraints

All values in input are integers.

1 \leq N \leq 2 \times 10^5

0 \leq K \leq 10^{18}

1 \leq A_i \leq 10^6\ (1 \leq i \leq N)

1 \leq F_i \leq 10^6\ (1 \leq i \leq N)

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N

Output

Print the minimum possible score of the team.

Sample Input 1

3 5
4 2 1
2 3 1

Sample Output 1

2

They can achieve the score of 2, as follows:

Member 1 does 4 sets of training and eats Food 2 in (4-4) \times 3 = 0 seconds.

Member 2 does 1 set of training and eats Food 3 in (2-1) \times 1 = 1 second.

Member 3 does 0 sets of training and eats Food 1 in (1-0) \times 2 = 2 seconds.

They cannot achieve a score of less than 2, so the answer is 2.

Sample Input 2

3 8
4 2 1
2 3 1

Sample Output 2

0

They can choose not to do exactly K sets of training.

Sample Input 3

11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2

Sample Output 3

12

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdint.h>
#include <stdlib.h>

#include <algorithm>
#include <iostream>
#include <numeric>
#include <vector>
using namespace std;
using default_counter_t = int64_t;

// macro
#define let auto const&
#define overload4(a, b, c, d, name, ...) name
#define rep1(n) \
  for (default_counter_t i = 0, end_i = default_counter_t(n); i < end_i; ++i)
#define rep2(i, n)                                                           \
  for (default_counter_t i = 0, end_##i = default_counter_t(n); i < end_##i; \
       ++i)
#define rep3(i, a, b)                                    \
  for (default_counter_t i       = default_counter_t(a), \
                         end_##i = default_counter_t(b); \
       i < end_##i; ++i)
#define rep4(i, a, b, c)                                 \
  for (default_counter_t i       = default_counter_t(a), \
                         end_##i = default_counter_t(b); \
       i < end_##i; i += default_counter_t(c))
#define rep(...) overload4(__VA_ARGS__, rep4, rep3, rep2, rep1)(__VA_ARGS__)

#define rrep1(n) \
  for (default_counter_t i = default_counter_t(n) - 1; i >= 0; --i)
#define rrep2(i, n) \
  for (default_counter_t i = default_counter_t(n) - 1; i >= 0; --i)
#define rrep3(i, a, b)                                         \
  for (default_counter_t i         = default_counter_t(b) - 1, \
                         begin_##i = default_counter_t(a);     \
       i >= begin_##i; --i)
#define rrep4(i, a, b, c)                                          \
  for (default_counter_t                                           \
           i = (default_counter_t(b) - default_counter_t(a) - 1) / \
                   default_counter_t(c) * default_counter_t(c) +   \
               default_counter_t(a),                               \
           begin_##i = default_counter_t(a);                       \
       i >= begin_##i; i -= default_counter_t(c))
#define rrep(...) \
  overload4(__VA_ARGS__, rrep4, rrep3, rrep2, rrep1)(__VA_ARGS__)

#define ALL(f, c, ...)                                           \
  (([&](decltype((c)) cccc) {                                    \
    return (f)(std::begin(cccc), std::end(cccc), ##__VA_ARGS__); \
  })(c))

// function
template <class C>
constexpr C& Sort(C& a) {
  std::sort(std::begin(a), std::end(a));
  return a;
}
template <class C>
constexpr auto& Min(C const& a) {
  return *std::min_element(std::begin(a), std::end(a));
}
template <class C>
constexpr auto& Max(C const& a) {
  return *std::max_element(std::begin(a), std::end(a));
}
template <class C>
constexpr auto Total(C const& c) {
  return std::accumulate(std::begin(c), std::end(c), C(0));
}
template <typename T>
auto CumSum(std::vector<T> const& v) {
  std::vector<T> a(v.size() + 1, T(0));
  for (std::size_t i = 0; i < a.size() - size_t(1); ++i) a[i + 1] = a[i] + v[i];
  return a;
}
template <typename T>
constexpr bool ChMax(T& a, T const& b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
constexpr bool ChMin(T& a, T const& b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}

void In(void) { return; }

template <typename First, typename... Rest>
void In(First& first, Rest&... rest) {
  cin >> first;
  In(rest...);
  return;
}

template <class T, typename I>
void VectorIn(vector<T>& v, const I n) {
  v.resize(size_t(n));
  rep(i, v.size()) cin >> v[i];
}

void Out(void) {
  cout << "\n";
  return;
}

template <typename First, typename... Rest>
void Out(First first, Rest... rest) {
  cout << first << " ";
  Out(rest...);
  return;
}

constexpr auto yes(const bool c) { return c ? "yes" : "no"; }
constexpr auto Yes(const bool c) { return c ? "Yes" : "No"; }
constexpr auto YES(const bool c) { return c ? "YES" : "NO"; }

#ifdef USE_STACK_TRACE_LOGGER
#ifdef __clang__
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Weverything"
#include <glog/logging.h>
#pragma clang diagnostic pop
#endif  //__clang__
#endif  // USE_STACK_TRACE_LOGGER

int64_t N, K;
vector<int64_t> A, F;

template <typename T>
bool C(T x) {
  int64_t K_ = K;
  rep(i, N) {
    int64_t tmp = A[i] * F[i];
    if (tmp - K_ * F[i] > x) {
      return false;
    }
    int64_t c = max<int64_t>(0, (tmp - x + F[i] - 1) / F[i]);
    K_ -= c;
  }
  return true;
}

template <typename T>
T LowerBound(T s, T t, bool (*comp)(T x) = C) {
  T lb = s - 1, ub = t;

  while (ub - lb > 1) {
    T mid                    = (ub + lb) / 2;  // ub lb がともに大きい時オーバフローするので注意
    ((*comp)(mid) ? ub : lb) = mid;
  }

  return ub;  // ub = t のときはすべてfalse
}

signed main(int argc, char* argv[]) {
  (void)argc;
#ifdef USE_STACK_TRACE_LOGGER
  google::InitGoogleLogging(argv[0]);
  google::InstallFailureSignalHandler();
#else
  (void)argv;
#endif  // USE_STACK_TRACE_LOGGER
  In(N, K);

  A.resize(N);
  F.resize(K);
  rep(i, N) In(A[i]);
  rep(i, N) In(F[i]);

  Sort(A);
  ALL(sort, F, greater<int64_t>());

  cout << LowerBound<int64_t>(0, 1e15);

  return EXIT_SUCCESS;
}