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: Time Limit Exceeded)

#include <algorithm>
#include <iostream>
#include <utility>
#include <vector>

using namespace std;

int solve(int n, int m, vector<pair<int, int>>& mp, int idx, int cur) {
    if (idx == static_cast<int>(mp.size())) {
        return (m / cur >= n ? cur : 0);
    }

    int r = 0;
    for (int i = 0, c = 1; i <= mp[idx].second; i++, c *= mp[idx].first) {
        r = max(r, solve(n, m, mp, idx + 1, cur * c));
    }

    return r;
}

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

    vector<pair<int, int>> mp(0);
    for (int i = 2, mt = m; mt > 1; i++) {
        if (mt % i == 0) {
            mp.push_back(make_pair(i, 0));
            while (mt % i == 0) {
                mp[mp.size() - 1].second++;
                mt /= i;
            }
        }
    }

    cout << solve(n, m, mp, 0, 1) << endl;
}