#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 = { 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61 };
vector<vll> PP(P.size()-2);
bool isOK(ll index, ll key) {

    ll k = 1;
    ll D = sqrt(index);
    D = max(0ll, D - 3);
    rep(i, 8) {
        if (D * D <= index) {
            D++;
        }
        else {
            D--;
            break;
        }
    }
    k += D-1;

    D = cbrt(index);
    D -= 3;
    rep(i, 8) {
        if (D * D*D <= index) {
            D++;
        }
        else {
            D--;
            break;
        }
    }
    k += D-1;

    rep(p, PP.size()) {
        auto D = lower_bound(all(PP[p]),index)-PP[p].begin();
        
        k += (D);
    }
    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() {
    rep(i, P.size() - 2) {
        ll p = P[i + 2];
        ll k = 2;
        while (1) {
            ll d = k;
            bool c = true;
            rep(u, p-1) {
                if (d <= 2e18 / k) {
                    d *= k;
                }
                else {
                    
                    c = false;
                    break;
                }
            }
            if (c) {
                PP[i].push_back(d);
            }
            else {
                PP[i].push_back(2e18);
                break;
            }
            k++;
            d = 0;
        }
    }
	ll T;
	cin >> T;
	rep(t, T) {
        ll N;
        cin >> N;
        cout << binary_search(N) << endl;
	}
}