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>
using namespace std;
///////////////////////////////////////////
const long long int INF = INT64_MAX;
const long long int Mod = 1000000007;
using ll = long long int; using ci = const int;
using vi = vector<int>;
using Vi = vector<long long int>;
typedef pair<int, int> P; typedef pair<ll, ll>PLL;
#define pb(x) push_back(x)
#define mp(x,y) make_pair(x,y)
#define all(x) (x).begin(),(x).end()
#define rep(i,N) for(ll i=0;i<N;i++)
auto max(auto x,auto y){if(x>y){return x;}else{return y;}}
auto min(auto x,auto y){if(x>y){return y;}else{return x;}}
template<class T>bool chmax(T &former, const T &b) { if (former<=b) { former=b; return true; } return false; }
template<class T>bool chmin(T &former, const T &b) { if (b<=former) { former=b; return true; } return false; }
template<class T>T sqar(T x){ return x*x; }//sqrt(x)は平方根;
#define Sort(v) std::sort(v.begin(), v.end(), std::greater<decltype(v[0])>()) //降順でVをソート
typedef vector<vector<ll> > matrix;
void princ(auto x){cout<<x<<" ";}; void print(auto x){cout<<x<<"\n";};
///////////////////////////////////////////////////////////////////////////////////

ll n,m;

int main(){
    cin.tie(0);ios::sync_with_stdio(false);
    cin>>n>>m;
    ll ans = 1;
    if(n==1){
        ans = m;
    }else{
        for(ll i =1;i<m;i++){
            if((i*i)>m)break;
            if(m%i == 0){
                if(i <= m/n){
                    ans = i;
                }else{
                    break;
                }
            }
        }
    }
    print(ans);
    return 0;    
}