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: Wrong Answer)

/*
 * atcoder/abc112/d.cpp
 */

// C++ 14
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring> // memset
#include <cassert>
#include <map>
using namespace std;
#define ll long long
#define loop(__x, __start, __end) for(int __x = __start; __x < __end; __x++)
template <class T> ostream & operator << (ostream & out, vector<T> const & v) {
  for (auto &&a: v) out << a << " "; out << endl; return out;
}
template <class T> void dump(T &a) { cout << a << endl; }
template <class T> bool chmin(T &a, T b) { if (a > b) {a = b; return true; } return false; }
template <class T> bool chmax(T &a, T b) { if (a < b) {a = b; return true; } return false; }

ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }

int N; ll M;
void solve() {
  cin >> N >> M;
  vector<ll> D;
  for (ll i=1; i*i<=M; i++) if (M%i==0) D.push_back(i);

  ll max_ = 0;
  for (auto&& d: D) {
    if (N * d <= M) chmax(max_, d);
  }
  cout << max_ << endl;
}
int main() {
  // cout.precision(15); cout << fixed;

  solve();

  return 0;
}