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 <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>
using namespace std;
using namespace boost::multiprecision;

// Define
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using int128 = __int128;
using cint = cpp_int;
const ll MOD = 1e9 + 7;
const ll INF = LONG_MAX;
const ull MAX = ULONG_MAX;
#define endl '\n'
#define space ' '
#define def inline auto
#define func inline constexpr ll
#define run __attribute__((constructor)) def _
#define all(v) begin(v), end(v)

// Debug
#define debug(...)                                                             \
    {                                                                          \
        cerr << __LINE__ << ": " << #__VA_ARGS__ << " = ";                     \
        for (auto &&X : {__VA_ARGS__}) cerr << "[" << X << "] ";               \
        cerr << endl;                                                          \
    }

// Loop
#define inc(i, a, n) for (ll i = (a), _##i = (n); i <= _##i; ++i)
#define dec(i, a, n) for (ll i = (a), _##i = (n); i >= _##i; --i)
#define each(i, a) for (auto &&i : a)
#define rep(i, n) inc(i, 0, n - 1)

// Stream
#define input(a) scanf("%lld", &(a))
#define output(a) printf("%lld\n", (a))
#define fout(n) cout << fixed << setprecision(n)
#define fasten cin.tie(0), ios::sync_with_stdio(0)

// Speed
run() { fasten, fout(10); }
#pragma GCC optimize("-O3")
#pragma GCC target("avx")

// Math
func gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
func lcm(ll a, ll b) { return a * b / gcd(a, b); }

signed main() {
    ll N, M, ANS = 0;
    cin >> N >> M;
    inc(i, 1, sqrt(M)) {
        if (M % i == 0 && M / i >= N) {
            ANS = max(ANS, i);
        }
        if (M % i == 0 && i >= N) {
            ANS = max(ANS, M / i);
        }
        debug(i, M / i);
    }
    cout << ANS << endl;
}

// for compilation: g++ -O3 -o _ _.cpp -std=c++17