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)

/***************************************************
 * Problem Name : D.cpp
 * Problem Link :
 * Contest Name :
 * OJ           :
 * Verdict      : Trying
 * Date         : 2018-10-06
 * Problem Type :
 * Author Name  : Saikat Sharma
 * University   : CSE, MBSTU
 ***************************************************/
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<climits>
#include<cstring>
#include<string>
#include<sstream>
#include<cmath>
#include<vector>
#include<queue>
#include<cstdlib>
#include<deque>
#include<stack>
#include<map>
#include<set>
#define __FastIO ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define SET(a,v) memset(a,v,sizeof(a))
#define pii pair<int,int>
#define pll pair <ll, ll>
#define debug cout <<"#########\n";
#define nl cout << "\n";
#define sp cout << " ";
#define sl(n) scanf("%lld", &n)
#define sf(n) scanf("%lf", &n)
#define si(n) scanf("%d", &n)
#define ss(n) scanf("%s", n)
#define pf(n) scanf("%d", n)
#define pfl(n) scanf("%lld", n)
#define all(v) v.begin(), v.end()
#define Pow2(x) ((x)*(x))
#define Mod(x, m) ((((x) % (m)) + (m)) % (m))
#define Max3(a, b, c) max(a, max(b, c))
#define Min3(a, b, c) min(a, min(b, c))
#define pb push_back
#define mk make_pair
#define MAX 100005
#define INF 1000000000
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

template <typename T>
std::string NumberToString ( T Number ) {
    std::ostringstream ss;
    ss << Number;
    return ss.str();
}
ll lcm (ll a, ll b) {
    return a * b / __gcd (a, b);
}
/************************************ Code Start Here ******************************************************/
int main () {
    //~ __FastIO;
    ll n, m;
    cin >> n >> m;

    if (m % n == 0) {
        cout << m / n << "\n";

    } else {
		cout << m%n << "\n";
    }

    return 0;
}