Problem:
Score : 400 points

Problem Statement

You are given integers N and M.

Consider a sequence a of length N consisting of positive integers such that a_1 + a_2 + ... + a_N = M. Find the maximum possible value of the greatest common divisor of a_1, a_2, ..., a_N.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

N \leq M \leq 10^9

Input

Input is given from Standard Input in the following format:

N M

Output

Print the maximum possible value of the greatest common divisor of a sequence a_1, a_2, ..., a_N that satisfies the condition.

Sample Input 1

3 14

Sample Output 1

2

Consider the sequence (a_1, a_2, a_3) = (2, 4, 8). Their greatest common divisor is 2, and this is the maximum value.

Sample Input 2

10 123

Sample Output 2

3

Sample Input 3

100000 1000000000

Sample Output 3

10000

Solution in Ruby (2.7.1) (judge verdict: Time Limit Exceeded)

require "prime"
n, m = gets.chomp.split.map(&:to_i)

factors = Prime.prime_division(m)
#全素因数を求める
factors = factors.map do |factor|
  (factor[1] + 1).times.map do |i|
  factor[0] ** i
  end
end
until (factors.size == 1) do
  factors = factors.each_slice(2).map{ |a, b|
    if b
      a.product(b).map{|c| c.reduce(&:*)}
    else
      a
    end
  }
end

factors = factors[0].sort.reverse
puts factors.find{|fac| fac <= (m/n)}