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 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

using namespace std;

typedef long long int lli;
typedef pair<int, int> ii;
typedef pair<int, lli> il;

int main() {
    int N, M;
    cin >> N >> M;

    int sM = sqrt(M);
    vector<int> list;
    for (int i = 1; i <= sM; i++) {
        if (M % i == 0) {
            list.push_back(i);
            if (M/i != i) {
                list.push_back(M/i);
            }
        }
    }
    sort(list.begin(), list.end(),
        [](const int &a, const int &b) {
            return a > b;
        });
    
    for (vector<int>::iterator it = list.begin(); it != list.end(); ++it) {
        int m = *it;
        if (M / m >= N) {
            cout << m << endl;
            return 0;
        }
    }
}