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 t, n, x, y;
int tmp[3];
long long calc2(long long x, long long t) {
  if (t == 1) {
    return 0;
  }
  long long nt = t / 4;
  if (x < nt) {
    return calc2(x, nt);
  }
  if (x < nt * 2) {
    return calc2(x - nt, nt) + 2 * nt;
  }
  if (x < nt * 3) {
    return calc2(x - nt * 2, nt) + 3 * nt;
  }
  if (x < nt * 4) {
    return calc2(x - nt * 3, nt) + 1 * nt;
  }
}
long long calc3(long long x, long long t) {
  if (t == 1) {
    return 0;
  }
  long long nt = t / 4;
  if (x < nt) {
    return calc3(x, nt);
  }
  if (x < nt * 2) {
    return calc3(x - nt, nt) + 3 * nt;
  }
  if (x < nt * 3) {
    return calc3(x - nt * 2, nt) + 1 * nt;
  }
  if (x < nt * 4) {
    return calc3(x - nt * 3, nt) + 2 * nt;
  }
}
int main() {
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    x = (n - 1) / 3;
    y = (n - 1) % 3 + 1;
    long long t = 1, v = 1;
    while (v <= x) {
      x -= v;
      v *= 4;
      t++;
    }
    long long ans = y * v;
    if (y == 1) {
      ans += x;
    }
    if (y == 2) {
      ans += calc2(x, v);
    }
    if (y == 3) {
      ans += calc3(x, v);
    }
    printf("%lld\n", ans);
  }
  return 0;
}
