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 parse(string s) {
  long long k = 1, ret = 0;
  for (int i = 0; i < s.size(); i++)
    ret += k * (long long)(s[s.size() - i - 1] - '0'), k *= 4;
  return ret;
}
string change(string s) {
  string ret = s;
  for (int i = 0; i < ret.size(); i++) {
    if (ret[i] == '1')
      ret[i] = '2';
    else if (ret[i] == '2')
      ret[i] = '3';
    else if (ret[i] == '3')
      ret[i] = '1';
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long idx = (n + 2) / 3, m = n % 3;
    long long k = 0, sum = 0;
    while (sum < idx) sum += (1ll << (2 * k)), k++;
    k--;
    long long buf = (1ll << (2 * k)), c = sum - buf + 1;
    string ans = "1";
    for (int i = k - 1; i >= 0; i--) {
      long long div = (1ll << (2 * i));
      long long d = (idx - c) / div;
      ans += (char)('0' + d);
      c += div * d;
    }
    long long f = parse(ans);
    long long s = parse(change(ans));
    if (m == 1)
      cout << f << '\n';
    else if (m == 2)
      cout << s << '\n';
    else
      cout << (f ^ s) << '\n';
  }
  return 0;
}
