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: Runtime Error)

#include<stdio.h>

int main()
{
 long N;
 long M;
 long i;
 long factor=1;
 long divisors;
 long rem;
 long commonshares;
 long out;

 scanf("%ld %ld",&N,&M);
 if((M%2)==0)
 {
  factor=2;
 }else{
  for(i=3;i*i<M;i+=2)
  {
   if((M%i)==0)
   {
    factor=i;
    break;
   }
  }  
 }
 divisors=M/factor;
 commonshares=(divisors-(divisors%N))/N;
 rem=divisors%commonshares;
 if(rem==0)
 {
  out=factor*commonshares;
 }else{
  out=factor;
 }
 printf("%ld\n",out);
 
 return 0;
}