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 <iostream>
using namespace std;
#include <vector>

vector<int> v;

int dfs(int i, int m, int n){
    if (m < n) return 0;
    if (i <= 0) return 1;
    
    int ret = max(dfs(i-1,m,n),dfs(i-1,m/v[i-1],n) * v[i-1]);
    return ret;
}

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

    

    int a = 2;
    int m = M;
    while (m >= a * a) {
        if (m % a == 0) {
            m /= a;
            v.push_back(a);
        } else {
            a++;
        }
    }

    v.push_back(m);

    int ans = dfs(v.size(),M,N);

    cout << ans << endl;

    return 0;

}