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;
using LL = long long;
LL mat[3][4] = {
    {0, 1, 2, 3},
    {0, 2, 3, 1},
    {0, 3, 1, 2},
};
LL recalc(LL n, int typ) {
  LL ans = 0;
  for (int i = 0; i <= 31; i++) {
    if ((n & (1ll << i + i)) || (n & (1ll << i + i + 1))) {
      ans |= (mat[typ][(n >> i + i) & 3] << i + i);
    }
  }
  return ans;
}
LL calc(LL n, int typ) {
  LL m = 1;
  int len = 2;
  while (n > m) {
    n -= m;
    len += 2;
    m <<= 2;
  }
  LL ans = recalc(n - 1, typ);
  if (typ != 1) {
    ans |= 1ll << len - 2;
  }
  if (typ != 0) {
    ans |= 1ll << len - 1;
  }
  return ans;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    LL n;
    scanf("%I64d", &n);
    LL f = (n + 2) / 3;
    int typ = (n - 1) % 3;
    printf("%I64d\n", calc(f, typ));
  }
  return 0;
}
