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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

void divisor(vector<int> &d, int n)
{
    for (int i = 1; i * i <= n; i++)
    {
        if(n % i == 0)
        {
            d.push_back(i);
            if(i * i != n)d.push_back(n / i);
        }
    }
    
}

int main()
{
    int n,m;
    cin >> n >> m;

    vector<int> mm;
    divisor(mm, m);
    sort(mm.begin(), mm.end());
    //for (int i = 0; i < mm.size(); i++)cout << mm[i] << " ";cout << endl;

    int ans = mm[mm.size() - 1];
    for (int i = 0; i < mm.size(); i++)
    {
        if(mm[i] > n)
        {
            ans = mm[i - 1];
            break;
        }
    }

    
    cout << ans << endl;
}