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 Ruby (2.7.1) (judge verdict: Accepted)

# O(n * log(n)) で解く

def g(i, n, t) $a_sums[i] + h(i+1, n, t) end    # O(n)
def h(i, n, t) (i..n).inject(0) do |s,j| s + t / $fs[j] end end    # O(n)
def afs(i) $as[i] * $fs[i] end    # O(1)
def nlr(n, l, i) [n - i, l - $a_sums[i], $rs[i+1]] end    # O(1)

n, k = gets.split.map(&:to_i)
$as = gets.split.map(&:to_i)
$fs = gets.split.map(&:to_i)
 
l = $as.sum - k

if (l <= 0)
  p 0
  exit
end

# O(n)
max_f = $fs.max
min_a, max_a = l, 0
# O(n)
$as.each do |a|
  min_a = a if (a < min_a)
  max_a = a if (a > max_a)
end

ub_af = max_a * max_f
q = max_f + ub_af

# O(n)
r = h(0, n-1, q)
t = (l + n) * q / r

if (min_a > t)
  ub = t

  rr = r + n
  lb = (l * q + rr - 1)/ rr

  s = 0
  xs = []
  # O(n)
  # push の回数は 3 * n を超えないはず
  (0..n-1).each do |i|
    f = $fs[i]
    aa = ub / f
    s += aa
    (aa * f).step(lb, -f) do |x| xs.push(-x) end
  end
else
  # O(n * log(n))
  $as.push(0).sort!
  $fs.push(max_f).sort_by! {|f| -f}

  # O(n * log(n))
  $as, $fs = $as.zip($fs).sort_by do |a,f| a*f end.transpose

  r = 0
  # O(n)
  $a_sums = $as.map do |a| r += a end

  $rs = []
  # O(n)
  n.downto(1).inject(0) do |r,i| $rs[i] = r + q / $fs[i] end

  lb = ub = nil
  ilb = nil

  # O(n + log(n))
  m = (0..n).bsearch do |i|
    next false if (i == 0)
    next true if (i == n)

    nn, ll, r = nlr(n, l, i)
    af = afs(i)

    next false if (af * (r + nn) / q < ll)
    next true if ((af * r + q - 1) / q - nn >= ll)

    next false if (lb)
    next true if (ub)

    # ここは一度しか通らない

    # O(n)
    if (h(i+1, n, af) < ll)
      lb, ilb = af, i
      next false
    else
      ub = af
      next true
    end
  end

  # O(n)
  if (lb)
    af = afs(m-1)
    ub = af if (g(m-1, n, af) > l)
  elsif (ub)
    af = afs(m)
    lb, ilb = af, m if (g(m, n, af) < l)
  end

  # O(1)
  unless (lb && ub)
    ilb = m == 1 ? 1: m - 1

    nn, ll, r = nlr(n, l, m-1)

    lbm, ubm = afs(m-1), afs(m)

    ub = [ub || ub_af, (ll + nn) * q / r, ubm].min

    rr = r + nn
    lb = [lb || 0, (ll * q + rr - 1)/ rr, lbm].max
  end

  s = $a_sums[ilb-1]
  xs = []
  # O(n)
  # push の回数は 3 * n を超えないはず
  (ilb..n).each do |i|
    a, f = $as[i], $fs[i]
    aa = [ub / f, a].min
    s += aa
    (aa * f).step(lb, -f) do |x| xs.push(-x) end
  end
end

# O(n) のアルゴリズムもあるが、この問題のサイズでは組み込みの sort を使った方が速いようだ
# O(n * log(n))
p(- xs.sort[s - l])