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 d1[] = {0, 2, 3, 1};
long long fn1(long long start, long long pos, long long len) {
  long long zzz = len / 4;
  if (len == 1) return start;
  for (long long i = 0; i < 4; i++) {
    long long l = i * zzz;
    long long r = l + (zzz - 1);
    if (l <= pos && pos <= r) return fn1(start + d1[i] * zzz, pos - l, zzz);
  }
  return -1;
}
long long d2[] = {0, 3, 1, 2};
long long fn2(long long start, long long pos, long long len) {
  long long zzz = len / 4;
  if (len == 1) return start;
  for (long long i = 0; i < 4; i++) {
    long long l = i * zzz;
    long long r = l + (zzz - 1);
    if (l <= pos && pos <= r) return fn2(start + d2[i] * zzz, pos - l, zzz);
  }
  return -1;
}
long long solve() {
  long long n;
  cin >> n;
  n--;
  long long row = n / 3;
  long long col = n % 3;
  if (col == 0) {
    long long pos = row;
    long long temp = 1;
    long long start = 1;
    while (true) {
      if (pos - temp >= 0) {
        pos -= temp;
        temp *= 4;
        start *= 4;
        continue;
      }
      break;
    }
    cout << start + pos << "\n";
  }
  if (col == 1) {
    long long pos = row;
    long long temp = 1;
    long long start = 2;
    while (true) {
      if (pos - temp >= 0) {
        pos -= temp;
        temp *= 4;
        start *= 4;
        continue;
      }
      break;
    }
    cout << fn1(start, pos, temp) << "\n";
  }
  if (col == 2) {
    long long pos = row;
    long long temp = 1;
    long long start = 3;
    while (true) {
      if (pos - temp >= 0) {
        pos -= temp;
        temp *= 4;
        start *= 4;
        continue;
      }
      break;
    }
    cout << fn2(start, pos, temp) << "\n";
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  long long t = 1;
  cin >> t;
  while (t-- != 0) {
    long long stat = solve();
  }
  return 0;
}
