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 <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <numeric>
#include <climits>
#include <iterator>
#include <iomanip>
#include <stack>
#include <set>
#include <cstdio>
#include <bitset>
#include <functional>
#include <tuple>
#include <list>
#include <deque>
#include <sstream>
using namespace std;
const constexpr int INF = 1e9;

#define FOR(i, a, n) for (ll i = (ll)a; i<(ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)

typedef long long ll;
typedef vector<int> VI;
typedef pair<ll, ll> P;
struct Less{
   bool operator()(const pair<string, int>&a, const pair<string, int>&b){
        if(a.first==b.first){
            return a.second < b.second;
        }
       return a.first < b.first;
    }
};
const constexpr ll MOD = 1e9 + 7;


ll extGCD(ll a, ll b, ll& x, ll& y){
    if(b==0){
        x=1;
        y=0;
        return a;
    }
    ll d = extGCD(b, a%b, y, x);
    y -= a/b*x;
    return d;
}
ll GCD(ll a, ll b){
    if(b==0) return a;
    return GCD(b, a%b);
}


int main() {
    ll N, M;  cin >> N >> M;
    
    ll ans=0;
    for(int i=2; i*N<=M; ++i){
        if(M%i!=0) continue;
        else{
            ll t=M/i; //tより小さい Nこの数字でtを表せられるか
            if(t*(t-1)/2<t||N*(t-1)<t) continue;
            ans = i;
        }
    }
    cout << ans << endl;
    
	return 0;
}