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 long long read() {
  char c = getchar();
  long long x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return x;
}
long long get(long long x, long long y, long long o) {
  if (x == 1) return 1;
  long long t = x / 4;
  if (o == 2) {
    if (y <= t) return get(t, y, o);
    if (y <= 2 * t) return get(t, y - t, o) + 2 * t;
    if (y <= 3 * t) return get(t, y - t * 2, o) + 3 * t;
    return get(t, y - 3 * t, o) + t;
  } else {
    if (y <= t) return get(t, y, o);
    if (y <= 2 * t) return get(t, y - t, o) + 3 * t;
    if (y <= 3 * t) return get(t, y - t * 2, o) + t;
    return get(t, y - t * 3, o) + t * 2;
  }
}
inline long long work() {
  long long n = read(), now = 0, c = 0, op = n % 3;
  n = (n - 1) / 3 + 1;
  while (now + (1ll << c) < n) now += (1ll << c), c += 2;
  long long res = now * 3;
  if (op == 1) return res + n - now;
  if (op == 2) return res + (1ll << c) + get(1LL << c, n - now, 2);
  return res + (1ll << c) * 2 + get(1LL << c, n - now, 3);
}
signed main() {
  long long T = read();
  while (T--) printf("%lld\n", work());
  return 0;
}
