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 md = 998244353;
const int xn = 1e5 + 10;
const int xm = -20 + 10;
const int SQ = 450;
const int sq = 1e3 + 10;
const int inf = 1e9 + 10;
const long long INF = 1e18 + 10;
long long power(long long a, long long b) {
  return (!b ? 1
             : (b & 1 ? a * power(a * a % md, b / 2) % md
                      : power(a * a % md, b / 2) % md));
}
long long zarb(long long a, long long b) { return (a * b + 10 * md) % md; }
long long jaam(long long a, long long b) { return (a + b + 10 * md) % md; }
string abc = "abcdefghijklmnopqrstuvwxyz";
long long n, qq, k, ind, sum, m = 0;
int main() {
  cin >> qq;
  while (qq--) {
    cin >> n;
    n--;
    ind = n / 3 + 1;
    sum = 0;
    k = 0;
    while (true) {
      sum += (1LL << (2 * k));
      if (ind <= sum) break;
      ++k;
    }
    sum -= (1LL << (2 * k));
    long long a = ind - sum - 1 + (1LL << (2 * k));
    long long last = (1LL << (k + k + 1));
    long long dx = (ind - sum - 1) / 4;
    long long x = (ind - sum - 1);
    long long b = last;
    long long kk = 0;
    while (x) {
      long long t = x % 4;
      if (t % 4 < 3 && t % 4)
        ++t;
      else if (t % 4)
        t -= 2;
      b += t * (1LL << (2 * kk));
      ++kk;
      x /= 4;
    }
    long long c = a ^ b;
    if (n % 3 == 0)
      cout << a << '\n';
    else if (n % 3 == 1)
      cout << b << '\n';
    else
      cout << c << '\n';
  }
  return 0;
}
