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 mod1 = 19260817;
const long long mod2 = 19660813;
long long n;
long long cur;
long long R;
void solve() {
  --n;
  cur = n % 3;
  n = n / 3;
  while ((1LL << R) <= n) {
    n -= (1LL << R);
    R += (1 << 1);
  }
  for (long long i = 0; i < R; i += 2) {
    long long tmp = n >> i & 3;
    if (tmp > 0) {
      n ^= tmp << i;
      tmp = (tmp + cur) % 3;
      if (tmp == 0) tmp = 3;
      n = (n ^ (tmp << i));
    }
  }
  if (cur == 1)
    n |= 1LL << (R + 1);
  else if (cur == 0)
    n |= (1LL << R);
  else {
    n = n | (1LL << R);
    n = n | (1LL << (R + 1));
  }
  printf("%lld\n", n);
}
int main() {
  int cas;
  scanf("%d", &cas);
  while (cas--) {
    R = 0;
    scanf("%lld", &n);
    solve();
  }
  return 0;
}
