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 (Clang 3.8.0) (judge verdict: Accepted)

// vim: shiftwidth=2 expandtab foldmethod=marker

// include {{{
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <cmath>
#include <utility>
#include <cstdlib>
#include <random>
#include <chrono>
#include <limits>
// }}}

// debug(...) {{{
#ifdef LOCAL_DEBUG
#include "./debug.h"
#else
bool debug_enabled = false;
#define debug(...)
#define debugn(...)
#define debug_reset()
#endif

// }}}

// bin {{{
template<class T>
std::string bin(int bits, const T& x) {
  char buf[bits + 1];
  buf[bits] = '\0';
  for(int i = 0; i < bits; i++) {
    if((x >> i) & 1) {
      buf[bits - i - 1] = '1';
    } else {
      buf[bits - i - 1] = '0';
    }
  }
  return std::string(buf);
}
std::string bin(int x) {
  if(x < 0) return bin(32, x);
  for(int i = 4; i < 31; i++) {
    int y = 1 << i;
    if(x < y) return bin(i, x);
  }
  return bin(32, x);
}
// }}}

// misc {{{
using namespace std;
template<class T, class U>

using umap = unordered_map<T, U>;
template<class T>
using uset = unordered_set<T>;

typedef long long ll;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)

template<class T>
void sort(T& arr) { // {{{
  std::sort(arr.begin(), arr.end());
} // }}}
template<class T, class By>
void sort_by(T& arr, By by) { // {{{
  std::sort(arr.begin(), arr.end(), [by](auto l, auto r){return  by(l) < by(r); });
} // }}}

template<class T>
void reverse(T& arr) { // {{{
  std::reverse(arr.begin(), arr.end());
} // }}}

class Random { // {{{
  public:
  std::uniform_int_distribution<> uniform_int;
  std::mt19937 rnd;
  std::bernoulli_distribution bern_dist;
  int uniform(int l, int r) {
    return uniform_int(rnd, std::uniform_int_distribution<>::param_type(l, r));
  }
  int next(int N) {
    return uniform(0, N - 1);
  }
  int bern(double p) {
    return bern_dist(rnd, std::bernoulli_distribution::param_type(p));
  }
}; // }}}

template <class A>
ostream& out_vec(const vector<A>& x, const string& sep) { // {{{
  rep(i, x.size()) {
    cout << x[i];
    if(i < x.size() - 1)
      cout << sep;
  }
  return cout;
} // }}}
// }}}

// modulo {{{
ll modpow(ll a, ll n, ll mod) {
    ll res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

ll modplus(ll a, ll b, ll mod) {
  ll x = ((a % mod) + (b % mod)) % mod;
  if(x < 0) x += mod;
  return x;
}

ll modminus(ll a, ll b, ll mod) {
  return modplus(a, -b, mod);
}

// }}}

class UnionFind { // {{{
  vector<int> parent;

  public:

  UnionFind(int N): parent(N) {
    rep(i, N) parent[i] = i;
  }

  int root(int x) {
    if(parent[x] == x) return x;
    else return parent[x] = root(parent[x]);
  }

  void unite(int x, int y) {
    int rx = root(x);
    int ry = root(y);
    if(rx != ry) {
      parent[rx] = ry;
    }
  }
}; // }}}

// main {{{
void solve();
int main() {
  cout << setprecision(20);
  solve();
}
// }}}

void solve() {
  ll N, K;
  cin >> N >> K;
  vector<ll> A(N);
  rep(i, N) cin >> A[i];
  vector<ll> F(N);
  rep(i, N) cin >> F[i];

  sort(A.begin(), A.end());
  sort(F.begin(), F.end());

  auto ok = [&](ll x) {
    ll k = 0;
    rep(i, N) {
      k += max(0ll, A[i] - x / F[N - i - 1]);
    }
    return k <= K;
  };

  ll r = A[N - 1] * F[N - 1];
  ll l = -1;

  while(l + 1 < r) {
    ll mid = (l + r) / 2;
    debugn(mid, ok(mid));
    if(ok(mid)) r = mid;
    else l = mid;
  }
  ll ans = r;

  cout << ans << endl;
}