Consider the infinite sequence s of positive integers, created by repeating the following steps:

  1. Find the lexicographically smallest triple of positive integers (a, b, c) such that 
    * a ⊕ b ⊕ c = 0, where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). 
    * a, b, c are not in s. 
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2]. 
  2. Append a, b, c to s in this order. 
  3. Go back to the first step. 



You have integer n. Find the n-th element of s.

You have to answer t independent test cases.

A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.

Each of the next t lines contains a single integer n (1≤ n ≤ 10^{16}) — the position of the element you want to know.

Output

In each of the t lines, output the answer to the corresponding test case.

Example

Input


9
1
2
3
4
5
6
7
8
9


Output


1
2
3
4
8
12
5
10
15

Note

The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ... 
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const int INF = 1e9;
const long long IINF = 1e18;
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const char dir[4] = {'D', 'R', 'U', 'L'};
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &x : v) is >> x;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (int i = 0; i < v.size(); ++i) {
    os << v[i] << (i + 1 == v.size() ? "" : " ");
  }
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  cout << '(' << p.first << ',' << p.second << ')';
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const map<T, U> &m) {
  os << '{';
  for (auto itr = m.begin(); itr != m.end(); ++itr) {
    os << '(' << itr->first << ',' << itr->second << ')';
    if (++itr != m.end()) os << ',';
    --itr;
  }
  os << '}';
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &s) {
  os << '{';
  for (auto itr = s.begin(); itr != s.end(); ++itr) {
    os << *itr;
    if (++itr != s.end()) os << ',';
    --itr;
  }
  os << '}';
  return os;
}
void debug_out() { cerr << '\n'; }
template <class Head, class... Tail>
void debug_out(Head &&head, Tail &&...tail) {
  cerr << head;
  if (sizeof...(Tail) > 0) cerr << ", ";
  debug_out(move(tail)...);
}
template <typename T>
T gcd(T x, T y) {
  return y != 0 ? gcd(y, x % y) : x;
}
template <typename T>
T lcm(T x, T y) {
  return x / gcd(x, y) * y;
}
template <class T1, class T2>
inline bool chmin(T1 &a, T2 b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T1, class T2>
inline bool chmax(T1 &a, T2 b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
int verify(int n) {
  int MAX = 10000;
  vector<int> check(MAX, 0);
  vector<int> res;
  for (int i = 1; i < MAX; ++i) {
    if (!check[i]) {
      for (int j = i + 1; j < MAX; ++j) {
        if ((i ^ j) < MAX && !check[j] && !check[i ^ j]) {
          res.emplace_back(i);
          res.emplace_back(j);
          res.emplace_back(i ^ j);
          check[i] = check[j] = check[i ^ j] = 1;
          break;
        }
      }
    }
  }
  return res[n];
}
long long n;
vector<long long> b = {0, 2, 3, 1};
pair<long long, long long> dfs(long long v) {
  if (v == 0) return {4, 8};
  if (v == 1) return {5, 10};
  if (v == 2) return {6, 11};
  if (v == 3) return {7, 9};
  pair<long long, long long> p = dfs(v / 4 - 1);
  p.first <<= 2LL;
  p.second <<= 2LL;
  p.first |= v % 4;
  p.second |= b[v % 4];
  return p;
}
void solve() {
  if (n < 3) {
    cout << n + 1 << '\n';
    return;
  }
  long long g = n / 3 - 1;
  pair<long long, long long> p = dfs(g);
  vector<long long> ans = {p.first, p.second, (p.first ^ p.second)};
  cout << ans[n % 3] << '\n';
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  for (; t--;) {
    cin >> n;
    --n;
    solve();
  }
}
