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;
inline bool setmin(int &x, int y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(int &x, int y) { return (y > x) ? x = y, 1 : 0; }
inline bool setmin(long long &x, long long y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(long long &x, long long y) { return (y > x) ? x = y, 1 : 0; }
const int N = 100000;
const int inf = (int)1e9 + 1;
const long long big = (long long)1e18 + 1;
const int P = 239;
const int P1 = 31;
const int P2 = 57;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const int MOD2 = 998244353;
const long double eps = 1e-9;
const double pi = atan2(0, -1);
const int ABC = 26;
string bin_str(int x) {
  string res;
  while (x) {
    res += char('0' + x % 2);
    x /= 2;
  }
  reverse(res.begin(), res.end());
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  cout << fixed;
  int T;
  cin >> T;
  while (T--) {
    long long x;
    cin >> x;
    x--;
    long long t = x % 3;
    x /= 3;
    for (long long b = 0;; b += 2) {
      if ((1LL << b) <= x) {
        x -= (1LL << b);
        continue;
      }
      long long ans = 0;
      long long pw = 0;
      while (x > 0) {
        long long v = x % 4;
        if (v == 0) {
          v = 0;
        } else {
          v += t;
          if (v > 3) {
            v -= 3;
          }
        }
        ans += (1LL << pw) * v;
        pw += 2;
        x /= 4;
      }
      ans += (t + 1) * (1LL << b);
      cout << ans << "\n";
      break;
    }
  }
  return 0;
}
