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 <iostream>
#include <vector>
#include <algorithm>
#include <cstdlib>
using namespace std;
typedef long long int ll;
int main(){
  ll n,m;
  cin >> n>>m;
  vector <ll>p;
for(ll i=1;i*i<=m;i++) if(m%i==0) p.push_back(i);
ll ma=-1;
for(int i=0;i<p.size();i++){
  if((m/(m/p[i]))>=n) ma=max(ma,(m/p[i]));
}
for(int i=0;i<p.size();i++){
if((m/p[i])>=n) ma=max(ma,p[i]);
}
cout <<ma<<endl;
}