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;
long long n;
long long x, y, z, w;
void init() {
  x = 0;
  y = 0;
  z = 0;
  w = 0;
}
long long sol(long long bs, long long i, long long z) {
  if (z == 0) return bs;
  long long t = (1ll << (z - 2));
  long long t2 = (i - 1) / t;
  if (t2 == 0) return sol(bs, i, z - 2);
  if (t2 == 1) return sol(bs + t * 2, (i - 1) % t + 1, z - 2);
  if (t2 == 2) return sol(bs + t * 2 + t, (i - 1) % t + 1, z - 2);
  if (t2 == 3) return sol(bs + t, (i - 1) % t + 1, z - 2);
}
long long sol1(long long n) {
  init();
  x = (n - 1) / 3 + 1;
  for (long long i = 0;; i += 2) {
    y += (1ll << i);
    if (y >= x) {
      return (1ll << (i + 1)) - 1 - (y - x);
    }
  }
}
long long sol2(long long n) {
  init();
  x = (n - 2) / 3 + 1;
  for (long long i = 0;; i += 2) {
    y += (1ll << i);
    if (y >= x) {
      z = (1ll << i) - (y - x);
      return sol((1ll << (i + 1)), z, i);
    }
  }
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    scanf("%lld", &n);
    x = 0;
    y = 0;
    z = 0;
    w = 0;
    if (n % 3 == 1) {
      printf("%lld\n", sol1(n));
    } else if (n % 3 == 2) {
      printf("%lld\n", sol2(n));
    } else {
      printf("%lld\n", (sol1(n - 2) ^ sol2(n - 1)));
    }
  }
  return 0;
}
