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;
inline int read() {
  int v = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    v = v * 10 + c - '0';
    c = getchar();
  }
  return v * f;
}
int H;
void mian(int _ = -1) {
  long long n;
  scanf("%lld", &n);
  if (n <= 3) {
    printf("%lld\n", n);
    return;
  }
  long long pos = (n - 1) / 3 + 1;
  H = n - pos * 3 + 3;
  int w = 0;
  while (1) {
    if (pos <= (1ll << w)) {
      break;
    }
    pos -= 1ll << w;
    w += 2;
  }
  long long st = 1ll << w;
  if (H == 1) {
    printf("%lld\n", st + pos - 1);
    return;
  }
  long long A = 0;
  pos--;
  for (int i = w - 2; i >= 0; i -= 2) {
    long long CW = 1ll << i;
    long long NW = 1ll << (i + 2);
    long long tmp = pos % NW / CW;
    if (tmp == 0)
      ;
    else if (tmp == 1)
      A |= 2ll << i;
    else if (tmp == 2)
      A |= 3ll << i;
    else
      A |= 1ll << i;
  }
  A += 2ll << w;
  if (H == 2) {
    printf("%lld\n", A);
    return;
  }
  if (H == 3) {
    printf("%lld\n", A ^ (st + pos));
    return;
  }
}
int main() {
  int t;
  scanf("%d", &t);
  for (int i = 1; i <= t; i++) {
    mian();
  }
}
