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 (1.42.0) (judge verdict: Accepted)

use std::io::Read;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

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

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();
    let m: usize = iterator.next().unwrap().parse().unwrap();

    let divisors = enumerate_divisors(m);

    let mut ans = 0;
    for d in divisors {
        if d > m / n {
            break;
        }
        if (m - (n - 1) * d) % d == 0 {
            ans = d;
        }
    }
    return ans.to_string();
}

fn enumerate_divisors(n: usize) -> Vec<usize> {
    let mut divisors = Vec::new();

    let mut i = 1;
    while i * i <= n {
        if n % i == 0 {
            divisors.push(i);
            if n / i != i {
                divisors.push(n / i);
            }
        }
        i += 1;
    }
    divisors.sort();
    return divisors;
}