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 <stack>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <cmath>
#include <climits>
#include <ctime>
#include <fstream>
#include <functional>
#define endl '\n'
#define ll long long
#define mod 1000000007
#define NCONS 200001
//#define NCONS 100001
//#define NCONS 1001
//#define NCONS 101
#define MCONS 2001
#define LIMIT 1000000000
#define TRUE 1
#define FALSE 0
#define toRadian(degree) ((degree) * (M_PI / 180.))
#define toDegree(radian) ((radian) * (180. / M_PI))
using namespace std;
struct Point{int x; int y;};
struct PPoint{Point x; Point y;};
ll gcd(ll a, ll b) { if(a < b) swap(a, b); if(b <= 0) return a; return gcd(b, a % b); }

int main(void)
{
    cin.tie(NULL); ios_base::sync_with_stdio(false);
    ll N, M; cin >> N >> M;
    ll gccd = 1;
    vector<ll> factors;
    for(ll i = 2; i * i <= M; i++)
    {
        if(M % i != 0) continue;
        factors.push_back(i);
        if(i * i != M)
            factors.push_back(M / i);
    }
    
    for(auto factor : factors)
    {
        if(M - (N-1) * factor <= 0) continue;
        if(M >= N * factor)
            gccd = max(gccd, factor);
    }
   
    cout << gccd << endl;
    return 0;
}



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