## solution

$x = {p_1}^{k_1} {p_2}^{k_2} \dots {p_n}^{k_n}$と素因数分解できたとき、 $xy = z^2$ for some $z$となるような最小の$y = \Pi \{ p_i \mid k_i \text{ is odd} \}$である。

## implementation

#include <cstdio>
#include <vector>
#include <set>
#include <cmath>
typedef long long ll;
using namespace std;

vector<int> sieve_of_eratosthenes(int n) { // enumerate primes in [2,n] with O(n log log n)
vector<bool> is_prime(n+1, true);
is_prime[0] = is_prime[1] = false;
for (int i = 2; i*i <= n; ++i)
if (is_prime[i])
for (int k = i+i; k <= n; k += i)
is_prime[k] = false;
vector<int> primes;
for (int i = 2; i <= n; ++i)
if (is_prime[i])
primes.push_back(i);
return primes;
}
vector<ll> factors(ll n, vector<int> const & primes) {
vector<ll> result;
for (int p : primes) {
if (n < p *(ll) p) break;
while (n % p == 0) {
result.push_back(p);
n /= p;
}
}
if (n != 1) result.push_back(n);
return result;
}

int main() {
ll n; scanf("%lld", &n);
auto primes = sieve_of_eratosthenes(sqrt(n));
vector<ll> ps = factors(n, primes);
set<ll> qs;
for (ll p : ps) {
if (qs.count(p)) {
qs.erase(p);
} else {
qs.insert(p);
}
}
ll r = 1;
for (ll q : qs) r *= q;
printf("%lld\n", r);
return 0;
}