#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
#define FOR(i,m,n) for(int i=(m);i<(n);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
using ll = long long;
constexpr int INF = 0x3f3f3f3f;
constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr double EPS = 1e-8;
constexpr int MOD = 1000000007;
// constexpr int MOD = 998244353;
constexpr int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
constexpr int dy8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dx8[] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U> inline bool chmax(T &a, U b) { return a < b ? (a = b, true) : false; }
template <typename T, typename U> inline bool chmin(T &a, U b) { return a > b ? (a = b, true) : false; }
struct IOSetup {
  IOSetup() {
    std::cin.tie(nullptr);
    std::ios_base::sync_with_stdio(false);
    std::cout << fixed << setprecision(20);
  }
} iosetup;

std::vector<int> prime_sieve(int n, bool get_only_prime) {
  std::vector<int> prime, smallest_prime_factor(n + 1);
  std::iota(smallest_prime_factor.begin(), smallest_prime_factor.end(), 0);
  for (int i = 2; i <= n; ++i) {
    if (smallest_prime_factor[i] == i) prime.emplace_back(i);
    for (int p : prime) {
      if (i * p > n || p > smallest_prime_factor[i]) break;
      smallest_prime_factor[i * p] = p;
    }
  }
  return get_only_prime ? prime : smallest_prime_factor;
}

int main() {
  constexpr int M = 31;
  const vector<int> prime = prime_sieve(M, true);
  const int m = prime.size();
  int t; cin >> t;
  while (t--) {
    ll x; cin >> x;
    ll tmp = x;
    vector<int> p(m, 0);
    REP(i, m) {
      while (tmp % prime[i] == 0) {
        tmp /= prime[i];
        ++p[i];
      }
    }
    int ans = INF;
    FOR(mul, 2, M + 1) {
      ll num = 1, den = 1;
      tmp = mul;
      REP(i, m) {
        den *= p[i] + 1;
        int ex = p[i];
        while (tmp % prime[i] == 0) {
          tmp /= prime[i];
          ++ex;
        }
        num *= ex + 1;
      }
      if (num % den == 0 && num / den == 2) chmin(ans, mul);
    }
    cout << x * ans << '\n';
  }
  return 0;
}