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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define FOR(n) for(int i=0;i<n;i++)
#define FORJ(n) for(int j=0;j<n;j++)
#define PRN(n) printf("%d\n",n)
#define PRF(n) printf("%lf\n",n)
#define PRL(n) printf("%lld\n",n)
#define PRS(s) printf("%s\n",s)
#define PRC(c) printf("%c",c)
#define mod 1000000007
typedef long long int ll;
int u(const void *a, const void *b){
    return *(ll*)a-*(ll*)b;
}
int d(const void *a, const void *b){
    return *(ll*)b-*(ll*)a;
}
int z(int a){
  if(a<1)a*=-1;
  return a;
}
int min(int a,int b){
  if(a>b)return b;
  return a;
}
int max(int a,int b){
  if(a>b)return a;
  return b;
}
int gcd(int a,int b){
  if(!b)return a;
  return gcd(b,a%b);
}
int main(void){
  ll n,m,ans=1;
  scanf("%lld %lld",&n,&m);
  for(ll i=2;i<=n;i++)if(m%i<1&&m/i>=n)ans=max(ans,i);
  PRL(ans);
  return 0;
}