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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<math.h>

int main(){
  long n = 0;
  long m = 0;
  long i = 0;
  long best = -1;
  scanf("%ld %ld",&n,&m);
  for(i=1;i<=sqrt(m)+1;i++){
    if(m%i==0){
      if(i >= n && ((best-n)>(i-n) ||best == -1)){
        best = i;
      }
      if(m/i >= n && ((best-n)>((m/i)-n) ||best == -1)){
        best = m/i;
      }
    }
  }
  printf("%ld",m/best);
  return 0;
}