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>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma comment(linker, "/STACK:2000000")
using namespace std;
using ll = long long;
using db = long double;
using ii = pair<int, int>;
const int N = 2e5 + 5, LG = 19, MOD = 1e9 + 7;
const int SQ = 320;
const long double EPS = 1e-7;
vector<int> adj[100005];
int cnt = 0;
bitset<1024> bt;
int ans = 0;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    ll n;
    cin >> n;
    for (ll i = 0;; i++) {
      if (n > 3 * (1ll << (i + i))) {
        n -= (1ll << (i + i)) * 3;
      } else {
        if (n % 3 == 1) {
          n = 1 + n / 3;
          cout << (1ll << (i + i)) + n - 1 << '\n';
        } else if (n % 3 == 2) {
          n = 1 + n / 3;
          ll offset = (1ll << (i + i + 1));
          ll ord[] = {0, 2, 3, 1};
          n--;
          int bits = i + i;
          for (int j = 0; j < bits; j += 2) {
            offset += ord[n % 4] << j;
            n >>= 2;
          }
          cout << offset << '\n';
        } else if (n % 3 == 0) {
          n /= 3;
          ll offset = 3 * (1ll << (i + i));
          n--;
          ll ord[] = {0, 3, 1, 2};
          int bits = i + i;
          for (int j = 0; j < bits; j += 2) {
            offset += ord[n % 4] << j;
            n >>= 2;
          }
          cout << offset << '\n';
        }
        break;
      }
    }
  }
  return 0;
}
