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)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <deque>
#include <queue>
#include <array>
#include <set>
#include <map>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <utility>
#include <tuple>
#include <functional>
#include <bitset>
#include <cstdint>
#include <cassert>

using namespace std;
using i64 = int64_t;
using i32 = int32_t;
template<class T, class U> void init_n(vector<T>& v, size_t n, U x) 
{ v = vector<T>(n, x); }
template<class T> void init_n(vector<T>& v, size_t n) { init_n(v, n, T()); }
template<class T> void read_n(vector<T>& v, size_t n, size_t o = 0) 
{ v = vector<T>(n+o); for (size_t i=o; i<n+o; ++i) cin >> v[i]; }
template<class T> void read_n(T a[], size_t n, size_t o = 0)
{ for (size_t i=o; i<n+o; ++i) cin >> a[i]; }
template<class T> T gabs(const T& x) { return max(x, -x); }
#define abs gabs

i64 n, m;
int main() {
  cin >> n >> m;
  // m % x == 0 and m / x >= n
  i64 ans = 1;
  for (i64 x = 2; x <= min(m, i64(40000)); ++x) {
    if (m % x == 0) {
      i64 y = m / x;
      if (y >= n) ans = max(ans, x);
      if (x >= n) ans = max(ans, y);
    }
  }
  cout << ans << endl;
  return 0;
}