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 <bits/stdc++.h>
#include <type_traits>
using namespace std;

using ll = long long;

#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rrep(i, n) for (int i = 1; i <= (n); ++i)
#define drep(i, n) for (int i = (n)-1; i >= 0; --i)
#define ddrep(i, n) for (int i = n; i > 0; --i)
#define srep(i, s, t) for (int i = s; i < t; ++i)
#define ssrep(i, s, t) for (int i = s; i <= t; ++i)

#define rng(a) a.begin(), a.end()
#define pb push_back
#define eb emplace_back
#define fi first
#define se second

#define chmax(x, y) (x = max(x, y))
#define chmin(x, y) (x = min(x, y))

using pi = pair<int, int>;
using vi = vector<int>;
using vvi = vector<vi>;
using ld = long double;

template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}

template <class T> ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  rep(i, (int)v.size()) {
    if (i)
      os << ",";
    os << v[i];
  }
  os << "}";
  return os;
}

template <typename T, size_t S> void printArray(const T (&array)[S]) {
  for (auto val : array)
    std::cout << val << ", ";
  std::cout << "\n";
}

const int mod = 1e9 + 7;
const int inf = 1e9 + 5;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << std::setprecision(10);

  int n, m; std::cin >> n >> m;
  int ans = 1;
  for (int i = 2; i * i <= m; i++) {
    if (m % i == 0 && m / n >= i) ans = i;
  }
  std::cout << ans << "\n";
}