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: Time Limit Exceeded)

#include <iostream>
#include <fstream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <limits>
#include <queue>
#include <iomanip>
#include <set>

template<typename T> bool chmax(T &a,T b){
    if(a<b){
        a=b;
        return true;
    }
    return false;
}

template<typename T> bool chmin(T &a,T b){
    if(a>b){
        a=b;
        return true;
    }
    return false;
}

using namespace std;
using ll = long long int;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vvll> vvvll;
const int MOD=1000000007;
//const int MOD=998244353;
const int INTMAX=1001001000;
const ll LLMAX=1010010010010010000;


int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    ll N,M;
    cin>>N>>M;
    
    for(ll i=M;i>=1;i--){
        if(M%i==0 && M/i>=N){
            cout<<i<<endl;
            return 0;
        }
    }
    
    return 0;
}