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: Accepted)

n,m = gets.strip.split.map(&:to_i)
#約数列挙
require 'prime'
def _divisors(primes, k)
  Enumerator.new do |y|
    if primes.size == k
      y << 1
    else
      p, e = primes[k]
      _divisors(primes, k + 1).each {|d|
        (e + 1).times {|e1|
          y << p ** e1 * d
        }
      }
    end
  end
end
def divisors(n)
  _divisors(Prime.prime_division(n), 0).to_a.sort
end
ary = divisors(m)
k = 0
for i in 0..ary.size-1
  if ary[i] <= m.to_f/n
    k += 1
  end
end
puts ary[k-1]