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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

// #undef DEBUG  // Uncomment this line to forcefully disable debug print.
#if DEBUG
template <typename T>
void debug(T value) {
  std::cerr << value;
}
template <typename T, typename... Ts>
void debug(T value, Ts... args) {
  std::cerr << value << ", ";
  debug(args...);
}
#define dbg(...)                              \
  do {                                        \
    cerr << #__VA_ARGS__ << ": ";             \
    debug(__VA_ARGS__);                       \
    cerr << " (L" << __LINE__ << ")" << endl; \
  } while (0)
#else
#define dbg(...)
#endif

void read_from_cin() {}
template <typename T, typename... Ts>
void read_from_cin(T& value, Ts&... args) {
  std::cin >> value;
  read_from_cin(args...);
}
#define in(type, ...) \
  type __VA_ARGS__;   \
  read_from_cin(__VA_ARGS__);

template <typename T>
void write_to_cout(const T& value) {
  std::cout << value << std::endl;
}
template <typename T, typename... Ts>
void write_to_cout(const T& value, const Ts&... args) {
  std::cout << value << ' ';
  write_to_cout(args...);
}
#define out(...) write_to_cout(__VA_ARGS__);

#define all(x) (x).begin(), (x).end()
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)

using ll = long long;

using namespace std;

template <typename T>
std::map<T, int> Factorize(T x) {
  std::map<T, int> m;
  for (T i = 2; i * i <= x;) {
    if (x % i == 0) {
      m[i]++;
      x /= i;
    } else {
      ++i;
    }
  }
  if (x > 1) m[x]++;
  return m;
}

int main() {
  in(int, n, m);
  map<int, int> fs = Factorize(m);
  int ans = 1;
  function<void(map<int, int>::iterator, int)> dfs =
      [&](map<int, int>::iterator it, int x) {
        if (it == fs.end()) {
          if (m / x >= n) ans = max(ans, x);
          return;
        }
        rep(i, it->second + 1) {
          dfs(next(it), x);
          x *= it->first;
        }
      };
  dfs(fs.begin(), 1);
  out(ans);
}