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>
#define FOR(i, a, b) for(int i=(int)(a); i!=(int)(b); i++)
#define RFOR(i, a, b) for(int i=(int)(a); i!=(int)(b); i--)
#define REP(i, n) FOR(i, 0, n)
using namespace std;
using ll = long long;

vector<ll> enum_div(ll n)//nの約数を列挙
{
	vector<ll> ret;
	for(int 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(){
  ll n, m; cin >> n >> m;

  vector<ll> y = enum_div(m);

  ll ans = 1;
  for(auto x: y){
    if(m >= n*x) ans = x;
  }

  cout << ans << endl;

  return 0;
}