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 <iostream>
#include <vector>
#include <algorithm>
#include <iomanip>
#include <string>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <tuple>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cassert>
#include <cstdint>
#include <cctype>
#include <numeric>
#include <bitset>
#include <functional>

using namespace std;

using ll =  long long;
using Pll = pair<ll, ll>;
using Pii = pair<int, int>;

constexpr int INF = 1 << 30;
constexpr ll LINF = 1LL << 60;
constexpr ll MOD = 1000000007;
constexpr long double EPS = 1e-10;
constexpr int dyx[4][2] = {
    { 0, 1}, {-1, 0}, {0,-1}, {1, 0}
};

vector<ll> get_divisors(ll n) {
    vector<ll> divisors;
    for(ll i=1;i*i<=n;++i) {
        if(n % i) continue;
        divisors.push_back(i);
        if(i * i != n) divisors.push_back(n/i);
    }
    return divisors;
}

int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    ll n, m, ans = 1;
    cin >> n >> m;
    if(n == 1) {
        cout << m << endl;
        return 0;
    }
    
    vector<ll> divisors = get_divisors(m);
    for(ll d: divisors) {
        if(m/d >= n) {
            ans = max(d, ans);
        }
    }

    cout << ans << endl;
}