#include <bits/stdc++.h>
//#include<atcoder/all>
using namespace std;
//using namespace atcoder;
using ll = long long;
#define all(A) A.begin(),A.end()
using vll = vector<ll>;
#define rep(i, n) for (long long i = 0; i < (long long)(n); i++)
using Graph = vector<vector<ll>>;
vll P = {5,7,11,13,17,19,23,29,31,37};
vector<vll> PP(P.size());
set<ll> S;
vll PPP;
bool isOK(ll index, ll key) {

    ll k = 0;
    ll D = sqrt(index);
    D = max(0ll, D - 5);
    rep(i, 11) {
        if (D * D <= index) {
            D++;
        }
        else {
            D--;
            break;
        }
    }
    k += D;
    D = cbrt(index);
    D = max(0ll, D - 5);
    rep(i, 11) {
        if (D * D*D <= index) {
            D++;
        }
        else {
            D--;
            break;
        }
    }
    k += D-1;
    
   auto Ds = upper_bound(all(PPP),index)-PPP.begin();

   k += (Ds);
    
    return (k >= key);


}

// 汎用的な二分探索のテンプレ
ll binary_search(ll key) {
    ll ng = 0; //「index = 0」が条件を満たすこともあるので、初期値は -1
    ll ok = key * key+3; // 「index = a.size()-1」が条件を満たさないこともあるので、初期値は a.size()

    /* ok と ng のどちらが大きいかわからないことを考慮 */
    while (abs(ok - ng) > 1) {
        ll mid = (ok + ng) / 2;
        cout << "";
        if (isOK(mid, key)) ok = mid;
        else ng = mid;
    }
    return ok;
}

int main() {
    //set<ll> S;
    //S.insert(2e18);
    PPP.push_back(1e18+10);
    rep(i, P.size()) {
        ll p = P[i];
        ll k = 2;
        while (1) {
            ll d = k;
            bool c = true;
            rep(u, p - 1) {
                if (d <= (1e18+10) / k) {
                    d *= k;
                }
                else {

                    c = false;
                    break;
                }
            }
            if (c) {
                ll D = sqrt(d);
                bool ks = true;
                rep(i, 11) {
                    if (D - 5 + i <= 0)continue;
                    if ((D - 5 + i) * (D - 5 + i) == d)ks = false;
                    if (d % (D - 5 + i) == 0 && (D - 5 + i) * (D - 5 + i) == d / (D - 5 + i))ks = false;
                }
                
                if (ks&&!S.count(d)) {
                    S.insert(d);
                    PPP.push_back(d);
                }
            }
            else {
                //PP[i].push_back(2e18);
                break;
            }
            k++;
            d = 0;
        }
    }
    sort(all(PPP));
    ll T;
    cin >> T;
    rep(t, T) {
        ll N;
        cin >> N;
        cout << binary_search(N) << endl;
    }
}