#pragma GCC optimization ("O3")

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using vec = vector<ll>;
using mat = vector<vec>;
using pll = pair<ll,ll>;
using dvec = vector<double>;
using dmat = vector<dvec>;

#define INF (1LL<<61)
//#define MOD 1000000007LL 
#define MOD 998244353LL

#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<class T> inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;}
template<class T> inline int chmax(T& a, T b) {if(a<b) {a=b; return 1;} return 0;}

class mint {
public:
    ll x;
    mint(ll x=0) : x((x%MOD+MOD)%MOD) {}
    mint operator-() const {return mint(-x);}
    mint& operator+=(const mint& a) {if((x+=a.x)>=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<<a.x;}
using mvec = vector<mint>;
using mmat = vector<mvec>;

ll gcd(ll a, ll b)
{
    return !b?a:gcd(b,a%b);
}

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 nroot(ll a, ll n)
{
    ll ok = 0, ng = INF;
    while(abs(ok-ng) > 1) {
        ll x = (ok+ng)/2;
        (npow(x, n)<=a?ok:ng) = x; 
    }
    return ok;
}

ll log2(ll a)
{
    ll ret = 0, x = 1;
    while(x < a) ++ret, x *= 2;
    return ret;
}

void solve()
{
    ll N;
    cin >> N;
    
    ll ans = N*N, M = log2(N);
    REP(i,1,M+1) {
        REP(j,1,M+1) {
            if(gcd(i, j) > 1) continue;
            ll m = max(i, j);
            ans += max(nroot(N, m)-1, 0LL)*(N/m);
        }
    }
    PR(ans);
}

int main()
{	                                                                                          ll T;
    cin >> T;
    while(T--) solve();

    return 0;
}

/*



*/