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;
const long long int maxn = 201001;
long long int get(long long int len, long long int pos) {
  if (len == 1) return 1;
  if (pos < len / 4) {
    return get(len / 4, pos);
  }
  if (pos >= (3 * len / 4)) {
    return (len / 4) + get(len / 4, pos - (3 * len / 4));
  }
  if (pos >= len / 4 && pos < len / 2) {
    return len / 2 + get(len / 4, pos - len / 4);
  }
  if (pos >= len / 2 && pos < (3 * len / 4)) {
    return 3 * (len / 4) + get(len / 4, pos - len / 2);
  }
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int cn = (n - 1) / 3;
    cn++;
    long long int f1 = 0, s1 = 0;
    long long int tek = 1;
    long long int tekp = 1;
    long long int num = 1;
    long long int ans = 0;
    while (num <= cn) {
      if (num + tekp > cn) {
        ans = tek + (cn - num);
        break;
      }
      num += tekp;
      tekp *= 4;
      tek = tekp;
    }
    f1 = ans;
    if (n % 3 == 1) {
      cout << ans << "\n";
      continue;
    }
    tek = 2;
    tekp = 1;
    num = 1;
    ans = 0;
    while (num <= cn) {
      if (num + tekp > cn) {
        ans = tek + get(tekp, cn - num);
        break;
      }
      num += tekp;
      tekp *= 4;
      tek *= 4;
    }
    s1 = ans - 1;
    if (n % 3 == 2) {
      cout << ans - 1 << "\n";
      continue;
    }
    if (n % 3 == 0) {
      cout << (f1 ^ s1) << "\n";
    }
  }
}
