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)

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn read_vec2<T: std::str::FromStr>(n: usize) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

fn read_col<T: std::str::FromStr>(n: usize) -> Vec<T> {
    (0..n).map(|_| read()).collect()
}

fn diff(x: i32, y: i32) -> i32 {
    if x > y {
        x - y
    } else {
        y - x
    }
}

use std::cmp::max;

fn main() {
    let nm: Vec<u64> = read_vec();
    let n = nm[0];
    let m = nm[1];

    let mut dv: Vec<u64> = vec![];

    for i in 1..(((m as f64).sqrt() as u64) + 1) {
        if m % i == 0 {
            dv.push(i);
        }
    }

    let l = dv.len();

    if dv[l - 1] * dv[l - 1] != m {
        let tmp = m / dv[l - 1];
        dv.push(tmp);
    }
    for i in (0..(l - 1)).rev() {
        let tmp = m / dv[i];
        dv.push(tmp);
    }

    for d in dv {
        if d >= n {
            println!("{}", m / d);
            break;
        }
    }
}