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 <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < n; ++i)
using ll = long long;
using namespace std;
const int INF = 1e9;

vector<ll> divisor(ll n) {
    vector<ll> ret;
    for (ll i = 1; i * i <= n; ++i) {
        if (n % i == 0) {
            ret.push_back(i);
            if (i * i != n)
                ret.push_back(n / 1);
        }
    }
    sort(ret.rbegin(), ret.rend());
    return ret;
}

int main() {
    ll N, M;
    cin >> N >> M;

    vector<ll> div = divisor(M);

    for (auto p : div) {
        ll now = p * N;
        if (M < now)
            continue;
        else {
            cout << p << endl;
            return 0;
        }
    }

    cout << 1 << endl;

    return 0;
}