#include<bits/stdc++.h>
#include<atcoder/all>
using namespace std;
using namespace atcoder;
#define rep(i,n)for (int i = 0; i < int(n); ++i)
#define rrep(i,n)for (int i = int(n)-1; i >= 0; --i)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
template<class T> void chmax(T& a, const T& b) {a = max(a, b);}
template<class T> void chmin(T& a, const T& b) {a = min(a, b);}
using ll = long long;
using P = pair<int,int>;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<ll>;
using VVL = vector<VL>;

constexpr ll MX = 1000000000'000000000;

bool is_square(ll x) {
    ll l = 1, r = 2;
    while(r * r <= x) l = r, r *= 2;
    while(r - l > 1) {
        ll c = (l + r) / 2;
        if (c * c <= x) l = c;
        else r = c;
    }
    return l * l == x;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    VL d;
    for(int b = 3; b < 60; b++) {
        for(ll a = 2;; a++) {
            long long ab = 1;
            bool ok = true;
            rep(_, b) {
                if (ab > MX / a) {
                    ok = false;
                    break;
                }
                ab *= a;
            }
            if (!ok) break;
            if (!is_square(ab)) d.push_back(ab);
        }
    }
    sort(all(d));
    d.erase(unique(all(d)), d.end());
    // cout << d.size() << endl;
    // return 0;
    // rep(i, 10) cout << d[i] << " ";
    // cout << endl;
    int tt;
    cin >> tt;
    while(tt--) {
        int k;
        cin >> k;
        ll l = 1, r = 1000000001;
        while(r - l > 1) {
            ll c = (l + r) / 2;
            // # of <= c^2
            ll cnt = c + (lower_bound(all(d), c * c) - d.begin());
            if (cnt <= k) l = c;
            else r = c;
        }
        auto it = lower_bound(all(d), l * l);
        ll cnt = l + (it - d.begin());
        int rest = k - cnt;
        // cout << l << ' ' << r << ' ' << rest << ' ' << (it - d.begin()) << endl;
        if (rest == 0) {
            cout << l * l << '\n';
        } else {
            cout << it[rest - 1] << '\n';
        }
    }
}