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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
typedef long long ll;
typedef long double ld;
#define rep(i,n) for(ll i=0;i<(n);i++)
#define repr(i,n) for(ll i=(n-1);i>=0;i--)
#define all(x) x.begin(),x.end()
#define br cout << "\n";
using namespace std;
const long long INF = 1e10;
const long long MOD = 1e9+7;
using Graph = vector<vector<ll>>;
using pll = pair<ll, ll>;
template<class T> inline bool chmin(T &a, T b) { if(a > b){ a = b; return true;} return false;}
template<class T> inline bool chmax(T &a, T b) { if(a < b){ a = b; return true;} return false;}

// 0 false, 1 true 
// string to int : -48
// a to A : -32
// ceil(a)  1.2->2.0
// c++17	g++ -std=c++17 a.cpp

int main() {
    std::cout << std::fixed << std::setprecision(15);
    ll n, m; cin >> n >> m;
    ll ans = 1;

    for(ll i = 1; i <= min(m/n, (ll)100000); i++){
        ll d = m / i;
        ll amari = m % i;
        if(amari != 0) continue;
        if(d >= n) ans = max(ans, i);
    }
    cout << ans << endl;
}