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 <bits/stdc++.h>
 
using namespace std;
 
vector <long long int> prime;
 
void eratos()
{
	bool isprime[31624];
	memset(isprime,true,sizeof(isprime));
 
	for(int i=2;i<=31623;i++)
	{
		if(isprime[i])
		{
			prime.push_back(i);
			for(int j=2*i;j<=31623;j+=i)
			{
				isprime[j] = false;
			}
		}
	}
}
 


int main(void)
{
	cin.tie(0);
	ios::sync_with_stdio(false);
 
	long long int n,m;
 
	cin >> n >> m;
 
 	eratos();
 
	long long int t = m;
 	long long int res = 1;

	for(int i=prime.size()-1;i>=0;i--)
	{
		if(t==1)
		{
			break;
		}
		if(t%prime[i]==0)
		{
			while(1)
			{
				if(t%prime[i]!=0)
				{
					break;
				}
				if(res*prime[i]*n<=m)
				{
					res*=prime[i];
				}
				t/=prime[i];
			}
		}
	}
 	
 	if(t!=1)
 	{
 		if(t*n<=m)
 		{
 			res = t;
 		}
 	}
 
 	
	if(m%n==0)
	{
		res = max(res,m/n);
	}
 
	for(long long int i=2;i<=sqrt(m);i++)
	{
		if(m%i==0 && i*n<=m)
		{
			long long int x = i;
			while(1)
			{
				if(i*x*n>m || m%(i*x)!=0)
				{
					break;
				}
				x*=i;
			}
			res = max(res,x);
		}
	}
 
	cout << res << '\n';
 
	return 0;
}