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 (2.3.3) (judge verdict: Time Limit Exceeded)

require 'prime'
N, M = gets.split.map(&:to_i)
pd = M.prime_division.flat_map { |(n, e)| [n] * e }.sort
if M % N == 0
  puts M / N
  exit 0
end

INF = 1 << 60
$min = INF
def dfs_min(pd, i, n)
  return INF if n >= $min
  if n >= N
    $min = n
    return n
  end
  return INF if i == pd.size
  [
    dfs_min(pd, i + 1, n),
    dfs_min(pd, i + 1, n * pd[i])
  ].min
end

N2 = M / N
def dfs_max(pd, i, n)
  return -1 if n > N2
  return n if i == pd.size
  [
    dfs_max(pd, i + 1, n),
    dfs_max(pd, i + 1, n * pd[i])
  ].max
end

if N > M / N
  div = dfs_min(pd, 0, 1)
  ans = M / div
  puts ans
else
  ans = dfs_max(pd, 0, 1)
  puts ans
end