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 <algorithm>
#include <fstream>
#include <vector>
#include <deque>
#include <assert.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <stdio.h>
#include <string.h>
#include <utility>
#include <math.h>
#include <bitset>
#include <iomanip>
#include <complex>
using namespace std;
#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define rep(i,b) rng(i,0,b)
#define gnr(i,a,b) for(int i=int(b)-1;i>=int(a);i--)
#define per(i,b) gnr(i,0,b)
#define ALL(x) (x).begin(),(x).end()
#define RALL(x) (x).rbegin(),(x).rend()
#define IDX(vec, element_iter) distance((vec).begin(), element_iter)
#define print(x) cout<<(x)<<'\n'
#define print_vec(v) rep(i,v.size()){if(i!=0)cout<<" ";cout<<v[i];}cout<<'\n'
#define pe(x) cout<<(x)<<" "
#define DEBUG(x) cout<<#x<<": "<<x<<'\n'
#define DEBUG_VEC(v) cout<<#v<<":";rep(i,v.size())cout<<" "<<v[i];cout<<'\n'
#define pb push_back
#define mp make_pair
#define MOD 1000000007 //10^9+7
#define fi first
#define sc second
using ll=long long;
using pi=pair<ll,ll>;
using vi=vector<int>;
using vll=vector<ll>;
using vstr=vector<string>;
int dx[4]={0,0,-1,1};
int dy[4]={-1,1,0,0};
void yes(bool c) { if (c)print("Yes"); else print("No"); };

    // 約数一覧を返します
vector<ll> devisors(ll a){
    vector<ll> ret;
    for(ll i=1; i*i<=a; i++){
        if(a%i!=0) continue;
        ret.push_back(i);
        ll another = a/i;
        if(i!=another) ret.push_back(another);
    }
    sort(ALL(ret), greater<ll>());//大きい順にソート
    return ret;
}

void solve() {
    ll n,m; cin >> n >> m;
    vll d = devisors(m);
    ll t = m/n;
    rep(i,d.size()){
        if(d[i]<=t) {
            print(d[i]);
            return;
        }
    }
}

int main() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
	cout<<fixed<<setprecision(20);
    //int q; cin >> q;
	//while (q--)
		solve();
    return 0;
}