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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    int bit = -1;
    for (int i = 0; i < 30; i++) {
      long long cnt = 3ll * (1ll << (i + i));
      if (cnt < n) {
        n -= cnt;
      } else {
        bit = i;
        break;
      }
    }
    assert(~bit);
    long long base = (1ll << (bit + bit));
    long long ans[] = {base, base * 2, base * 3};
    int pat[] = {1, 2, 3};
    for (int i = bit - 1; i >= 0; i--) {
      for (int j = 0; j < 4; j++) {
        long long cnt = (3ll << (i + i));
        if (cnt < n) {
          n -= cnt;
        } else {
          if (j == 0) {
            break;
          } else {
            int k = j - 1;
            int c = 0;
            do {
              ans[c] += (1ll * pat[k] * (1ll << (i + i)));
              k++;
              c++;
              k %= 3;
            } while (c < 3);
          }
          break;
        }
      }
    }
    cout << ans[n - 1] << "\n";
  }
}
