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: Accepted)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cassert>
#include <climits>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <fstream>

using namespace std;
#define TOSTRING(x) #x
#define SZ(x) (int)(x).size()
#define REP(i,n) for(int i=0;i<(n);i++)
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REPR(i,n) for(int i=(n)-1;i>=0;i--)
#define ALL(s) (s).begin(), (s).end()
#define so(V) sort(ALL(V))
#define rev(V) reverse(ALL(V))
#define uni(v) v.erase( unique(ALL(v)) , v.end());
#define PAU system("pause")

typedef long long unsigned int llu;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<bool> vb;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef vector<vb> vvb;
const double EPS = 1e-9;
const int MOD = 1e9 + 7;
const int INF = (1 << 28);
const double PI = acos(-1);

typedef pair<int, int> P;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N,M;
	cin >> N >> M;
	vector<P> div;
	for (int i = 1; i*i <= M; ++i)
	{
		if (M%i == 0)
		{
			div.push_back(P(M/i,i));
			if (i*i != M)
			{
				div.push_back(P(i,M / i));
			}
		}
	}
	so(div); rev(div);
	int ok = 0;
	int ng = SZ(div);
	while (ng - ok > 1) {
		int mid = (ok + ng) / 2;
		if (div[mid].first >= N)ok = mid;
		else ng = mid;
	}
	cout << div[ok].second << endl;
	PAU;
	return 0;
}