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 <cstdio>
#include <iostream>
#include <string>
#include <vector>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <typeinfo>
#include <numeric>
#include <functional>
#include <unordered_map>
#include <bitset>
#include <stack>


using namespace std;
using ll = long long;
using ull = unsigned long long;

const ll INF = 1e16;
const ll MOD = 1e9 + 7;

#define REP(i, n) for(ll i = 0; i < n; i++)
#define Yes(condition) if(condition) cout << "Yes" << '\n'; else{ cout << "No" << '\n'; }
#define YES(condition) if(condition) cout << "YES" << '\n'; else{ cout << "NO" << '\n'; }
#define yes(condition) if(condition) cout << "yes" << '\n'; else{ cout << "no" << '\n'; }



ll gcd(ll a, ll b){
    return b? gcd(b, a % b) : a;
}

int main() {
    ll n, m;
    cin >> n >> m;
    vector<ll> b;
    for(ll i = 1; i <= sqrt(m); i++){
        if(m % i == 0){
            b.push_back(i);
            b.push_back(m / i);
        }
    }
    sort(b.begin(), b.end());
    ll ans = 1;
    for(ll i = 0; i < b.size(); i++){
        if(b[i] * n <= m){
            ans = b[i];
        }
    }
    
    cout << ans << endl;
}