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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

using int32 = int_fast32_t;
using int64 = int_fast64_t;

const int32 INF = 1e9;
const int32 MOD = 1e9+7;
const int64 LLINF = 1e18;

#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define ANS(n) cout << (n) << endl
#define REP(i,n) for(int64 i=0;i<(n);++i)
#define FOR(i,a,b) for(int64 i=(a);i<(b);i++)
#define FORR(i,a,b) for(int64 i=(a);i>=(b);i--)
#define ALL(obj) (obj).begin(),(obj).end()
#define pii pair<int32,int32>
#define pll pair<int64,int64>
#define pb(a) push_back(a)
#define mp make_pair

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int32 n,m;
    cin >> n >> m;
    FOR(i,n,m+1){
        if(m % i == 0){
            ANS(m/i);
            return 0;
        }
    }
    return 0;
}