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 <bits/stdc++.h>
#include <iostream>
#include <vector>
#include <string>
#include <numeric>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <iomanip>
#include <queue>

#define rep(i,n) for(int i=0;i<n;i++)
#define dup(x,y) (((x)+(y)-1)/(y)) // dup * y >= x なる最小のdup.
using namespace std;
typedef long long ll;
using Graph = vector<vector<int>>;

int main() {
	ll N, M;
	cin >> N >> M;

	ll sumA = M;
	
	for(ll i=1; i*i<=M; i++) {
		if(M%i==0) {
			if(M/i >= N && sumA > M/i) {
				sumA = M/i;
			}
		}
	}

	cout << M/sumA << endl;

	return 0;
}