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;
set<long long> ss;
vector<long long> s;
tuple<long long, long long, long long> find() {
  tuple<long long, long long, long long> mn;
  for (long long a = 1; a <= 300; a++) {
    for (long long b = a + 1; b <= 300; b++) {
      for (long long c = b + 1; c <= 300; c++) {
        if (!ss.count(a) && !ss.count(b) && !ss.count(c) &&
            ((a ^ b) ^ c) == 0) {
          return {a, b, c};
        }
      }
    }
  }
}
long long find_a(long long n) {
  long long s = 1, pw = 0, li = 1, ri = 1;
  while (1) {
    if (li <= n && n <= ri) break;
    pw += 2;
    s = (1ll << pw);
    long long len = (ri - li + 1);
    li += (len);
    long long new_len = (1ll << (pw + 1)) - (1ll << pw);
    ri = li + new_len - 1;
  }
  return s + (n - li);
}
long long gett(long long len, long long n, long long li, long long pw) {
  if (len == 1) return 0;
  long long i = n - li;
  long long b_id = (i / (len / 4));
  long long buff = 0;
  if (b_id > 1) buff += (1ll << pw);
  if (b_id > 2) buff += (1ll << (pw + 2));
  i %= (len / 4);
  return buff + gett(len / 4, i + li - (len / 4), li - (len / 4), pw - 2);
}
long long find_b(long long n) {
  long long li = 1, ri = 1, len = 1, pw = -2;
  while (1) {
    if (li <= n && n <= ri) {
      return gett(len, n, li, pw);
    } else {
      len *= 4;
      li = ri + 1;
      ri = li + len - 1;
      pw += 2;
    }
  }
}
void PUNKS_NOT_DEAD() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long num = (n - 1) / 3 + 1;
    long long a = find_a(num);
    long long b = -(find_b(num) - a * 2);
    long long c = (a ^ b);
    if ((n - 1) % 3 == 0)
      cout << a << '\n';
    else if ((n - 1) % 3 == 1)
      cout << b << '\n';
    else
      cout << c << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  PUNKS_NOT_DEAD();
}
