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 OCaml (4.02.3) (judge verdict: Accepted)

let rec floor_sqrt acc acc_x_2_x_r sq_acc_minus_z r sq_r =
  if r = 0 then acc
  else
    let sq_acc_minus_z' = sq_acc_minus_z + acc_x_2_x_r + sq_r in
    ( if sq_acc_minus_z' <= 0 then
        floor_sqrt (acc + r) ((acc_x_2_x_r lsr 1) + sq_r) sq_acc_minus_z'
      else
        floor_sqrt acc (acc_x_2_x_r lsr 1) sq_acc_minus_z) (r lsr 1) (sq_r lsr 2)
let floor_sqrt z = floor_sqrt 0 0 (~-z) (1 lsl 30) (1 lsl 60)

let () = Scanf.scanf "%d %d" @@ fun n m ->
  Printf.printf "%d\n" @@
  if n = 1 then m
  else Array.fold_left (List.fold_left max) 1 @@
    Array.init (floor_sqrt m) @@ fun i ->
      if 0 < m mod (i + 1)
      then []
      else (if m / (i + 1) < n then [] else [i + 1]) @ (if (i + 1) < n then [] else [m / (i + 1)])