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;

template<typename T> vector<T> divisor(T n){
    vector<T> ret;
    for (T i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            ret.push_back(i);
            if (i * i != n) { ret.push_back(n / i); }
        }
    }
    sort(ret.begin(), ret.end());
    return ret;
}

int main(){
    long long N, M;
    cin >> N >> M;
    vector<long long> Divisor = divisor(M);
    long long ans = 0;
    for (long long i = 0; i < Divisor.size(); i++) {
        if (Divisor[i] <= N) ans = max(ans, Divisor[i]);
    }
    cout << ans << endl;
    return 0;
}