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 Rust (1.15.1) (judge verdict: Accepted)

#[allow(unused)]
use std::io::*;
#[allow(unused)]
use std::str::*;
#[allow(unused)]
use std::collections::*;

pub fn read_from<T: FromStr, R: Read>(s: &mut R) -> T {
    let s = s.bytes().map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    if s.is_empty() {
        if cfg!(local) {
            // eprintln!("reached to EOF");
        }
        std::process::exit(0)
    }
    s.parse::<T>().unwrap_or_else(|_| panic!("faild to parse {}", s))
}

pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let mut stdin = stdin.lock();
    read_from(&mut stdin)
}

pub const MOD: u64 = 1000000007;

fn main() {
    loop {
        let n: u64 = read();
        let m: u64 = read();

        let check = |d: u64| m / d >= n;

        let mut ans = 1;
        for i in 1..m + 1 {
            if i * i > m {
                break;
            }
            if m % i == 0 {
                let j = m / i;
                if check(i) && ans < i {
                    ans = i;
                }
                if check(j) && ans < j {
                    ans = j;
                }
            }
        }

        println!("{}", ans);
    }
}