output_description: Print the maximum possible value of the greatest common divisor of a sequence
  a_1, a_2, ..., a_N that satisfies the condition.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N M
attempt: ###############################################################################
  
  from bisect import bisect_left as binl
  from copy import copy, deepcopy
  
  
  def intin():
      input_tuple = input().split()
      if len(input_tuple) <= 1:
          return int(input_tuple[0])
      return tuple(map(int, input_tuple))
  
  
  def intina():
      return [int(i) for i in input().split()]
  
  
  def intinl(count):
      return [intin() for _ in range(count)]
  
  
  def modadd(x, y):
      global mod
      return (x + y) % mod
  
  
  def modmlt(x, y):
      global mod
      return (x * sy) % mod
  
  
  def lcm(x, y):
      while y != 0:
          z = x % y
          x = y
          y = z
      return x
  
  
  def make_linklist(xylist):
      linklist = {}
      for a, b in xylist:
          linklist.setdefault(a, [])
          linklist.setdefault(b, [])
          linklist[a].append(b)
          linklist[b].append(a)
      return linklist
  
  
  def calc_longest_distance(linklist, v=1):
      distance_list = {}
      distance_count = 0
      distance = 0
      vlist_previous = []
      vlist = [v]
      nodecount = len(linklist)
  
      while distance_count < nodecount:
          vlist_next = []
          for v in vlist:
              distance_list[v] = distance
              distance_count += 1
              vlist_next.extend(linklist[v])
          distance += 1
          vlist_to_del = vlist_previous
          vlist_previous = vlist
          vlist = list(set(vlist_next) - set(vlist_to_del))
  
      max_distance = -1
      max_v = None
      for v, distance in distance_list.items():
          if distance > max_distance:
              max_distance = distance
              max_v = v
  
      return (max_distance, max_v)
  
  
  def calc_tree_diameter(linklist, v=1):
      _, u = calc_longest_distance(linklist, v)
      distance, _ = calc_longest_distance(linklist, u)
      return distance
  
  
  ###############################################################################
  
  
  def get_factors(x):
      retlist = []
      for i in range(1, int(x**0.5) + 3):
          if x % i == 0:
              retlist.append(i)
              retlist.append(x // i)
      return retlist
  
  
  def main():
      n, m = intin()
  
      flist = get_factors(m)
      flist.sort()
  
      for f in flist:
          if f >= n:
              print(m // f)
              return
  
  
  if __name__ == "__main__":
      main()
  
problem_description: 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.
samples: [{"input": "3 14", "output": "2\n    \n\nConsider the sequence (a_1, a_2, a_3) = (2, 4, 8). Their greatest common\ndivisor is 2, and this is the maximum value.\n\n* * *"}, {"input": "10 123", "output": "3\n    \n\n* * *"}, {"input": "100000 1000000000", "output": "10000"}]