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;
long long n;
long long T;
long long wei(long long a) {
  long long t = 0;
  while (a) {
    t++;
    a >>= 1;
  }
  return t;
}
long long suan(long long now, long long xuyao) {
  if (xuyao == 1) return now;
  long long A = wei(now);
  long long ss = 0;
  for (long long i = A - 1; i >= 2; i -= 2) {
    long long a = (1ll << (i - 1)) & now, b = (1ll << (i - 2)) & now;
    if (!a && !b) continue;
    if (a && !b) ss += (1ll << (i - 1)) + (1ll << (i - 2));
    if (!a && b) ss += 1ll << (i - 1);
    if (a && b) ss += 1ll << (i - 2);
  }
  ss += 1ll << A;
  return xuyao == 2 ? ss : now ^ ss;
}
signed main() {
  cin >> T;
  while (T--) {
    cin >> n;
    long long gg = (n - 1) / 3 + 1;
    long long sum = 0;
    for (long long i = 1; i <= 63; i += 2) {
      sum += 1ll << (i - 1);
      if (sum >= gg) {
        sum -= 1ll << (i - 1);
        gg -= sum + 1;
        gg += 1ll << (i - 1);
        cout << suan(gg, (n - 1) % 3 + 1) << '\n';
        break;
      }
    }
  }
  return 0;
}
