#include using namespace std; // clang-format off //#include //using namespace atcoder; using ll = long long; using ld = long double; using pii = pair; using pll = pair; using pdd = pair; using vii = vector; using vll = vector; using vdd = vector; using vvii = vector>; using vvll = vector>; using vvdd = vector>; using vvvii = vector>>; using vvvll = vector>>; using vvvdd = vector>>; template using P = pair; template using V1 = vector; template using V2 = vector>; template using V3 = vector>>; template using pque = priority_queue, greater>; #define _overload3(_1, _2, _3, name, ...) name #define rep1(n) for (ll i = 0; i < (n); i++) #define rep2(i, n) for (ll i = 0; i < (n); i++) #define rep3(i, a, b) for (ll i = (a); i < (b); i++) #define rep(...) _overload3(__VA_ARGS__, rep3, rep2, rep1)(__VA_ARGS__) #define rrep1(n) for (ll i = (n) - 1; i >= 0; i--) #define rrep2(i, n) for (ll i = (n) - 1; i >= 0; i--) #define rrep3(i, a, b) for (ll i = (b) - 1; i >= (a); i--) #define rrep(...) _overload3(__VA_ARGS__, rrep3, rrep2, rrep1)(__VA_ARGS__) #define all(x) (x).begin(), (x).end() #define sz(x) ((int)(x).size()) #define fi first #define se second #define pb push_back #define endl '\n' #define popcnt(x) __builtin_popcountll(x) #define uniq(x) (x).erase(unique((x).begin(), (x).end()), (x).end()); #define IOS ios::sync_with_stdio(false); cin.tie(nullptr); const int inf = 1 << 30; const ll INF = 1ll << 60; const ld DINF = numeric_limits::infinity(); const ll mod = 1000000007; //const ll mod = 998244353; const ld EPS = 1e-9; const ld PI = 3.1415926535897932; const ll dx[8] = {1, 0, -1, 0, 1, -1, -1, 1}; const ll dy[8] = {0, 1, 0, -1, 1, 1, -1, -1}; template bool chmax(T &a, const T &b) { if (a bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; } template T min(const vector &x) { return *min_element(all(x)); } template T max(const vector &x) { return *max_element(all(x)); } template ostream &operator<<(ostream &os, const pair &p) { return os << "(" << p.fi << ", " << p.se << ")"; } template ostream &operator<<(ostream &os, const vector &v) { os << "[ "; for (auto &z : v) os << z << " "; os << "]"; return os; } #ifdef _DEBUG #define show(x) cout << #x << " = " << x << endl; #else #define show(x) #endif ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; } ll lcm(ll a, ll b) { return (a / gcd(a, b)) * b; } ll rem(ll a, ll b) { return (a % b + b) % b; } // clang-format on // using mint = modint998244353; ll modpow(ll a, ll N, ll mod) { ll res = 1; // 例えば3=101(2)なので、下位bitから順に1ならa倍する while (N) { if (N & 1) res = res * a % mod; a = a * a % mod; N >>= 1; } return res; } bool isRoot(ll x) { ll res = (ll)(sqrt(x)); return res * res == x; } void solve(vll &b3) { ll k; cin >> k; // x以下の累乗数がcnt(x)個ある // cnt(x) >= Kとなるxの最小値 // (lb, ub] ll ub = INF, lb = 0; while (ub - lb > 1) { ll mid = (ub + lb) / 2; ll cnt = upper_bound(all(b3), mid) - b3.begin(); cnt += (ll)(sqrt(mid)); if (cnt >= k) ub = mid; else lb = mid; } cout << ub << endl; // } int main() { vll b3; rep(a, 2, 1000001) { ll x = a * a * a; while (1) { if (!isRoot(x)) b3.pb(x); if (x > INF / a) break; x *= a; } } sort(all(b3)); uniq(b3); int t; cin >> t; while (t--) solve(b3); return 0; } /* メモ 二分探索 b>=3の場合でa^2の形で表せないものを前もって列挙する b>=3よりa<=10^6まで考えれば良いので全部列挙できる ある数xより小さい累乗数は b=2のものはsqrt(x)で、b>=3は全列挙したものから二分探索することで求められる */