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 C++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <vector>
#include <string>

int main() {
#ifdef WYXLOCAL
	freopen("tmp.in", "r", stdin);
	freopen("tmp.out", "w", stdout);
#endif // WYXLOCAL
	int n, m;
	scanf("%d%d", &n, &m);
	int mx = sqrt(m) + 1;
	int ans = 1;
	for (int i = 1; i <= mx; ++i) {
		if (m % i == 0 && m / i >= n) ans = i;
	}
	if (n == 1) printf("%d\n", m);
	else printf("%d\n", ans);
	return 0;
}