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;
const int N = 1e6 + 50;
long long T;
long long n, nn, a[4] = {0, 2, 3, 1}, b[4] = {0, 3, 1, 2};
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%lld", &n);
    long long x = 1;
    while (n > x * 3) n -= x * 3, x <<= 2;
    if (n % 3 == 1)
      printf("%lld\n", x + n / 3);
    else if (n % 3 == 2) {
      n /= 3;
      long long ans = x << 1;
      for (long long i = x >> 2; i; i >>= 2) {
        long long dat = n / i;
        ans |= a[dat] * i;
        n %= i;
      }
      cout << ans << endl;
    } else {
      n--;
      n /= 3;
      long long ans = x << 1 | x;
      for (long long i = x >> 2; i; i >>= 2) {
        long long dat = n / i;
        ans |= b[dat] * i;
        n %= i;
      }
      cout << ans << endl;
    }
  }
  return 0;
}
