#pragma region Macros #include using namespace std; template inline bool chmax(T &a, T b) { if(a < b) { a = b; return 1; } return 0; } template inline bool chmin(T &a, T b) { if(a > b) { a = b; return 1; } return 0; } #ifdef DEBUG template ostream &operator<<(ostream &os, const pair &p) { os << '(' << p.first << ',' << p.second << ')'; return os; } template ostream &operator<<(ostream &os, const vector &v) { os << '{'; for(int i = 0; i < (int)v.size(); i++) { if(i) { os << ','; } os << v[i]; } os << '}'; return os; } void debugg() { cerr << endl; } template void debugg(const T &x, const Args &...args) { cerr << " " << x; debugg(args...); } #define debug(...) \ cerr << __LINE__ << " [" << #__VA_ARGS__ << "]: ", debugg(__VA_ARGS__) #define dump(x) cerr << __LINE__ << " " << #x << " = " << (x) << endl #else #define debug(...) (void(0)) #define dump(x) (void(0)) #endif struct Setup { Setup() { cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(15); } } __Setup; using ll = long long; #define OVERLOAD3(_1, _2, _3, name, ...) name #define ALL(v) (v).begin(), (v).end() #define RALL(v) (v).rbegin(), (v).rend() #define REP1(i, n) for(int i = 0; i < int(n); i++) #define REP2(i, a, b) for(int i = (a); i < int(b); i++) #define REP(...) OVERLOAD3(__VA_ARGS__, REP2, REP1)(__VA_ARGS__) #define UNIQUE(v) sort(ALL(v)), (v).erase(unique(ALL(v)), (v).end()) #define REVERSE(v) reverse(ALL(v)) #define SZ(v) ((int)(v).size()) const int INF = 1 << 30; const ll LLINF = 1LL << 60; constexpr int MOD = 1000000007; constexpr int MOD2 = 998244353; const int dx[4] = {1, 0, -1, 0}; const int dy[4] = {0, 1, 0, -1}; void Case(int i) { cout << "Case #" << i << ": "; } int popcount(int x) { return __builtin_popcount(x); } ll popcount(ll x) { return __builtin_popcountll(x); } #pragma endregion Macros ll memo[10000000 + 1]; void solve() { ll N; cin >> N; auto rec = [&](auto self, ll n) -> ll { if(n == 1) return 1; if(memo[n]) return memo[n]; ll ret = n * (n + 1) / 2; ll l = 1; while(l <= n) { ll q = n / l; ll r = n / q + 1; if(q < n) ret -= self(self, q) * (r - l); l = r; } return (memo[n] = ret); }; ll cnt = rec(rec, N) - 1; ll ans = cnt + (N * (N-1) / 2 - cnt) * 2; cout << ans << '\n'; } int main() { int T = 1; cin >> T; while(T--) solve(); }