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: Accepted)

#include <bits/stdc++.h>
using namespace std;
/*{{{*/  //template
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
constexpr int INF = numeric_limits<int>::max()/2;
constexpr long long LINF = numeric_limits<long long>::max()/3;
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
#define sz(x) (int)(x).size()
#define debug(x) cerr<<#x<<":"<<x<<endl
#define debug2(x,y) cerr<<#x<<","<<#y":"<<x<<","<<y<<endl
//struct fin{ fin(){ cin.tie(0); ios::sync_with_stdio(false); } } fin_;
struct Double{ double d; explicit Double(double x) : d(x){} };
ostream& operator<<(ostream& os,const Double x){ os << fixed << setprecision(20) << x.d; return os; }
template<typename T> ostream& operator<<(ostream& os,const vector<T>& vec){ os << "["; for(const auto& v : vec){ os << v << ","; } os << "]"; return os; }
template<typename T,typename U> ostream& operator<<(ostream& os, const map<T,U>& mp){ os << "{"; for(auto& p : mp){ os << p << ","; } os << "}"; return os; }
template<typename T,typename U> ostream& operator<<(ostream& os,const pair<T,U>& p){ os << "(" << p.first << ","<< p.second <<")"; return os; }
template<typename T> ostream& operator<<(ostream& os,const set<T>& st){ os<<"{"; for(T v:st) os<<v<<","; os <<"}"; return os; }
template<typename T,typename U> inline void chmax(T &x,U y){ if(y>x) x = y; }
template<typename T,typename U> inline void chmin(T &x,U y){ if(y<x) x = y; }
using ll = long long;
using ull = unsigned long long;
using pii = std::pair<int,int>;
using vi = std::vector<int>;
typedef vector<vi> vvi;
ll gcd(ll a,ll b){ if(b==0) return a; else return gcd(b,a%b); }
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n - 1); }
//constexpr double eps = 1e-14; 
constexpr double eps = 1e-10; 
constexpr ll mod = 1e9+7;
const int dx[]={1,0,-1,0} ,dy[] = {0,1,0,-1};
/*}}}*/

int main(){
    ll N,M; cin >> N >> M;
    vector<ll> d;
    for(ll i=1;i*i<=M;i++){
        if(M%i==0){
            d.emplace_back(i);
            d.emplace_back(M/i);
        }
    }
    sort(all(d));
    reverse(all(d));
    for(auto v : d){
        ll num = M / v;
        if(num >= N){
            cout << v << endl;
            return 0;
        }
    }
}