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 r(i,n) for(int i=0;i<n;i++)
#define int long long
using namespace std;
typedef pair<int,int>P;
typedef pair<int,P>P2;
#define fi first
#define se second
int dx[]={0,1,0,-1};
int dy[]={-1,0,1,0};

signed main(){
	int n,m;
	cin>>n>>m;
	vector<int>v;
	for(int i=1;i*i<=m;i++){
		if(m%i)continue;
		v.push_back(i);
		if(i*i!=m)v.push_back(m/i);
	}

	sort(v.begin(),v.end(),greater<int>());

	for(int j=0;j<v.size();j++){
		int i=v[j];
		//cout<<i<<endl;
		if(i*(n-1)>=m)continue;
		int x=i*(n-1);
		int p=m-x;
		if(__gcd(p,i)==i){
			cout<<i<<endl;
			return 0;
		}
	}
}