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 MOD = (1e9) + 7;
const long long N = (1e6) + 5;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tc = 63;
  cin >> tc;
  for (long long ti = 1; ti <= tc; ti++) {
    long long n;
    cin >> n;
    long long s = 1LL, e = 3LL;
    while (1) {
      if (n >= s && n <= e) break;
      s *= 4LL;
      e++;
      e *= 4LL;
      e--;
    }
    n -= s;
    long long r = (n) / 3LL;
    long long siz = s;
    long long bs = 2LL * s, cs = 3LL * s, rem;
    while (siz > 1LL) {
      r = n / 3LL;
      siz /= 4LL;
      long long bno = r / siz;
      if (!bno) {
        s = s;
        bs = bs;
        cs = cs;
      } else if (bno == 1) {
        s = s + siz;
        bs = bs + (2LL * siz);
        cs = cs + (3LL * siz);
      } else if (bno == 2) {
        s = s + (2LL * siz);
        bs = bs + (3LL * siz);
        cs = cs + (siz);
      } else {
        s = s + (3LL * siz);
        bs = bs + (siz);
        cs = cs + (2LL * siz);
      }
      n -= (bno * siz * 3);
    }
    if (n % 3 == 0)
      cout << s;
    else if (n % 3 == 1)
      cout << bs;
    else
      cout << cs;
    cout << endl;
  }
  return 0;
}
