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 N = 1000;
long long used[N], a[N];
signed main() {
  srand(123);
  long long cur = 1, cnt = 0;
  while (cur < 256) {
    while (used[cur]) cur++;
    for (long long i = cur + 1; i < N; i++)
      if ((cur ^ i) < N && !used[i] && !used[cur ^ i]) {
        cnt++;
        a[cnt] = cur;
        cnt++;
        a[cnt] = i;
        cnt++;
        a[cnt] = (cur ^ i);
        used[cur] = used[i] = used[cur ^ i] = 1;
        break;
      }
  }
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long nn = n;
    if (n <= 3) {
      cout << n << "\n";
      continue;
    }
    long long x = 1, lg = 0;
    while (x * 4 <= n) x *= 4, lg++;
    n -= (x - 1);
    n--;
    long long id = n / 3;
    long long f[3];
    f[0] = id + x;
    f[1] = 2;
    while (lg > 0) {
      long long t = (1ll << (2 * lg - 2));
      if (id / t == 0) f[1] <<= 2;
      if (id / t == 1) f[1] <<= 2, f[1] |= 2;
      if (id / t == 2) f[1] <<= 2, f[1] |= 3;
      if (id / t == 3) f[1] <<= 2, f[1] |= 1;
      id %= t;
      lg--;
    }
    f[2] = (f[0] ^ f[1]);
    cout << f[n % 3] << "\n";
  }
  return 0;
}
