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 fi first
#define se second
#define show(x) cerr << #x << "=" << "\n"
typedef long long ll;
typedef unsigned long long ull;

using namespace std;

int n, m;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
#ifdef LOCAL_DEFINE
  FILE *stream1;
  //FILE *stream2;
  stream1 = freopen("in.txt", "r", stdin);
  //stream2 = freopen("out.txt", "w", stdout);
  if (stream1 == NULL) return 0;
  //if (stream2 == NULL) return 0;
#endif
  cin >> n >> m;
  int ans = 1;
  for (int i = 1; i * i <= m; i++) {
    if (m % i == 0) {
      if (i >= n) ans = max(ans, m / i);
      if (m / i >= n) ans = max(ans, i);
    }
  }
  cout << ans << endl;
#ifdef LOCAL_DEFINE
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s.\n";
  fclose(stream1);
  //fclose(stream2);
#endif
  return 0;
}