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)

open Printf open Scanf
let divisors v = 
	let rec f0 i a =
		if i*i > v then a
		else if i*i = v && v mod i = 0 then i::a
		else if v mod i = 0 then f0 (i+1) (i::(v/i)::a)
		else f0 (i+1) a in f0 1 []
let () = scanf " %d %d" @@ fun n m -> 
	let l = m/n in
	divisors m |> List.sort compare |> List.rev
	|> List.iter @@ fun v ->
		if v <= l then (printf "%d\n" v; exit 0)