//#include <tourist>
#include <bits/stdc++.h>
//#include <atcoder/all>
using namespace std;
//using namespace atcoder;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<ll, ll> p;
const int INF = 1e9;
const ll LINF = ll(1e18);
const int MOD = 1000000007;
const int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
const int Dx[8] = {0, 1, 1, 1, 0, -1, -1, -1}, Dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
#define yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define no cout << "No" << endl
#define NO cout << "NO" << endl
#define rep(i, n) for (int i = 0; i < n; i++)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define ALL(v) v.begin(), v.end()
#define debug(v)          \
    cout << #v << ":";    \
    for (auto x : v)      \
    {                     \
        cout << x << ' '; \
    }                     \
    cout << endl;
template <class T>
bool chmax(T &a, const T &b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
bool chmin(T &a, const T &b)
{
    if (b < a)
    {
        a = b;
        return 1;
    }
    return 0;
}
//cout<<fixed<<setprecision(15);有効数字15桁
//-std=c++14
//g++ yarudake.cpp -std=c++17 -I .
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; }
bool check(ll mid,ll k,ll n){
    //min^k<=nならtrue
    rep(i,k){
        n/=mid;
        if(n==0)return false;
    }
    return true;
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll t;
    cin >> t;
    rep(i, t)
    {
        //解説AC
        ll n;
        cin >> n;
        ll ans=n*n;//右と左が同じやつ
        ans+=n*(n-1);//右と左が違くて1^aのやつ
        for(ll i=1;i<32;i++){
            for(ll j=1;j<32;j++){
                if(gcd(i,j)!=1)continue;
                ll k=max(i,j);
                if(k==1)
                    continue; //右と左が同じやつになるので
                //xの候補をしぼります
                ll l=1,r=1e5;
                while(r-l>1){
                    ll mid=(r+l)/2;
                    if(check(mid,k,n)){
                        l=mid;
                    }
                    else{
                        r=mid;
                    }
                }
                ans+=(n/k)*(l-1);//(肩にのせる物の候補, k*h<=nになる物の数)*xの候補
                //(x^i)^(j*h)=(x^j)^(i*h)をやるiとjは互いに素
            }
        }
        cout<<ans<<"\n";
    }
}