#pragma GCC optimization ("O3") #include using namespace std; using ll = long long; using vec = vector; using mat = vector; using pll = pair; using dvec = vector; using dmat = vector; #define INF (1LL<<61) //#define MOD 1000000007LL #define MOD 998244353LL #define EPS (1e-10) #define PR(x) cout << (x) << endl #define PS(x) cout << (x) << " " #define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i)) #define FORE(i,v) for(auto (i):v) #define ALL(x) (x).begin(), (x).end() #define SZ(x) ((ll)(x).size()) #define REV(x) reverse(ALL((x))) #define ASC(x) sort(ALL((x))) #define DESC(x) {ASC((x)); REV((x));} #define BIT(s,i) (((s)>>(i))&1) #define pb push_back #define fi first #define se second template inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;} template inline int chmax(T& a, T b) {if(a=MOD) x-=MOD; return *this;} mint& operator-=(const mint& a) {if((x+=MOD-a.x)>=MOD) x-=MOD; return *this;} mint& operator*=(const mint& a) {(x*=a.x)%=MOD; return *this;} mint operator+(const mint& a) const {mint b(*this); return b+=a;} mint operator-(const mint& a) const {mint b(*this); return b-=a;} mint operator*(const mint& a) const {mint b(*this); return b*=a;} mint pow(ll t) const {if(!t) return 1; mint a=pow(t>>1); return (t&1?*this*a:a)*a;} mint inv() const {return pow(MOD-2);} mint& operator/=(const mint& a) {return *this*=a.inv();} mint operator/(const mint& a) const {mint b(*this); return b/=a;} }; istream &operator>>(istream& is, mint& a) {ll t; is>>t; a=t; return is;} ostream &operator<<(ostream& os, const mint& a) {return os<; using mmat = vector; ll npow(ll a, ll n) { ll ret = 1; REP(i,0,n) { if(ret >= INF/a) { ret = INF; break; } ret *= a; } return ret; } ll count_npow(ll x, ll n) { ll ok = 0, ng = INF; while(abs(ok-ng) > 1) { ll mid = (ok+ng)/2; if(npow(mid, n) <= x) ok = mid; else ng = mid; } return ok-1; } ll count_all_npow(ll x) { ll ret = 1; vec dp(60); for(ll i=59; i>=2; --i) { dp[i] = count_npow(x, i); for(ll j=i*2; j<=59; j+=i) dp[i] -= dp[j]; ret += dp[i]; } return ret; } int main() { ll T; cin >> T; while(T--) { ll K; cin >> K; ll ok = INF, ng = 0; while(abs(ok-ng) > 1) { ll mid = (ok+ng)/2; if(count_all_npow(mid) >= K) ok = mid; else ng = mid; } PR(ok); } return 0; } /* */