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<queue>
#include<cmath>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#define F(i,n,v) for(register int i=1;i<=n;i+=v)
#define R(i,n,v) for(register int i=n;i>=1;i-=v)
#define LL long long
#define mod 1000000007
#define inf 0x7ffffff
using namespace std;
inline int read();
inline int read()
{
	char c;
	while((c=getchar())<'0'||c>'9');
	int res=c-48;
	while((c=getchar())>='0'&&c<='9')
	   res=res*10+c-48;
	return res;
}const int MAX=10015;
bool prime(int x)
{
	for(int i=2;i<=sqrt(x);i++)
	if(x%i==0)return false;
	return true;
}int n,m,cnt,a[MAX];
int work()
{
	int p,x=m,ans=1;
	while(x>1)
	{
		for(int i=2;i<=x;i++)
		if(x%i==0){p=i;x/=i;break;}
		a[++cnt]=p;
	}sort(a+1,a+cnt);
	for(int i=1;i<=cnt;i++)
	{
		if(a[i]>n&&a[i-1]<n)continue;
		else ans*=a[i];
	}return ans;
}
int main()
{
	scanf("%d%d",&n,&m);
	if(prime(m)==true)printf("1");
	else 
	{
		if(m%n==0)printf("%d",m);
		else printf("%d",work());
	}
    return 0;
}