#include using namespace std; //using namespace atcoder; struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_; #define FOR(i, begin, end) for(int i=(begin);i<(end);i++) #define REP(i, n) FOR(i,0,n) #define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--) #define IREP(i, n) IFOR(i,0,n) #define Sort(v) sort(v.begin(), v.end()) #define Reverse(v) reverse(v.begin(), v.end()) #define all(v) v.begin(),v.end() #define SZ(v) ((int)v.size()) #define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x)) #define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x)) #define chmax(a, b) a = max(a, b) #define chmin(a, b) a = min(a, b) #define bit(n) (1LL<<(n)) #define debug(x) cout << #x << "=" << x << endl; #define vdebug(v) { cout << #v << "=" << endl; REP(i_debug, (int)v.size()){ cout << v[i_debug] << ","; } cout << endl; } #define mdebug(m) { cout << #m << "=" << endl; REP(i_debug, (int)m.size()){ REP(j_debug, (int)m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;} } #define pb push_back #define fi first #define se second #define int long long #define INF 1000000000000000000 template istream &operator>>(istream &is, vector &v){ for (auto &x : v) is >> x; return is; } template ostream &operator<<(ostream &os, vector &v){ for(int i = 0; i < (int)v.size(); i++) { cout << v[i]; if(i != (int)v.size() - 1) cout << endl; }; return os; } template ostream &operator<<(ostream &os, pair p){ cout << '(' << p.first << ',' << p.second << ')'; return os; } template void Out(T x) { cout << x << endl; } template void chOut(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); } using vec = vector; using mat = vector; using Pii = pair; using v_bool = vector; using v_Pii = vector; //int dx[4] = {1,0,-1,0}; //int dy[4] = {0,1,0,-1}; //char d[4] = {'D','R','U','L'}; //const int mod = 1000000007; const int mod = 998244353; int solve(int N){ N++; const int n = 62; int dp[n][2][3][2] = {}, dq[n][2][3][2] = {}; dp[n - 1][0][0][0] = 1; IREP(i, n - 1){ int d = (N >> i) & 1; REP(j, 2) REP(k, 3) REP(f, 2){ REP(x, 2){ if(j == 0 && d < x) continue; int j2 = j | (x < d); if(x == 0){ int f2 = (k == 2) | (f == 1 && k >= 1); dp[i][j2][0][f2] += dp[i + 1][j][k][f]; int c = k; if(k == 2 && f == 1) c = 1; dq[i][j2][0][f2] += dq[i + 1][j][k][f] + dp[i + 1][j][k][f] * c; }else{ dp[i][j2][min(k + 1, 2LL)][f] += dp[i + 1][j][k][f]; dq[i][j2][min(k + 1, 2LL)][f] += dq[i + 1][j][k][f]; } } } } int ans = 0; REP(k, 3) REP(f, 2){ int c = k; if(k == 2 && f == 1) c = 1; ans += dq[0][1][k][f] + dp[0][1][k][f] * c; } return ans; } signed main(){ int T; cin >> T; REP(_, T){ int N; cin >> N; int ans = solve(N); Out(ans); } return 0; }