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;
template <typename A, typename B>
string to_string(pair<A, B> p);
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);
string to_string(const string& s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (long long i = 0; i < static_cast<long long>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}
template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ")";
}
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
const long long INF = LONG_LONG_MAX / 10;
const double PI = 3.141592653589793238;
const long long MOD = 998244353;
bool many = true;
struct Solve {
  string endl = "\n";
  pair<long long, long long> Get(long long bit, long long idx) {
    if (bit == 0) {
      return pair<long long, long long>{1, 2};
    }
    long long some = 1ll << (bit - 2);
    auto q = Get(bit - 2, idx % some);
    q.first ^= (1ll << (bit - 2));
    q.second ^= (1ll << (bit - 1));
    long long cs = idx / some;
    if (cs == 1) {
      q.first ^= (1ll << (bit - 2));
      q.second ^= (1ll << (bit - 1));
    }
    if (cs == 2) {
      q.first ^= (1ll << (bit - 1));
      q.second ^= (1ll << (bit - 1)) ^ (1ll << (bit - 2));
    }
    if (cs == 3) {
      q.first ^= (1ll << (bit - 1)) ^ (1ll << (bit - 2));
      q.second ^= (1ll << (bit - 2));
    }
    q.first ^= 1ll << bit;
    q.second ^= 1ll << (bit + 1);
    return q;
  }
  Solve() {
    long long n;
    cin >> n;
    long long idx = (n - 1) / 3;
    long long pos = (n - 1) % 3;
    for (long long bit = 0;; bit += 2) {
      long long here = 1ll << bit;
      if (idx < here) {
        auto p = Get(bit, idx);
        cout << (pos == 0 ? p.first
                          : (pos == 1 ? p.second : (p.first ^ p.second)))
             << endl;
        break;
      }
      idx -= here;
    }
  }
};
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long CASES = 1;
  if (many) cin >> CASES;
  for (long long i = 1; i <= CASES; i++) {
    Solve me;
  }
}
