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;
class InputReader {
 public:
  InputReader() {}
  InputReader(const char *file_name) {
    input_file = fopen(file_name, "r");
    cursor = 0;
    fread(buffer, SIZE, 1, input_file);
  }
  inline InputReader &operator>>(int &n) {
    while (buffer[cursor] < '0' || buffer[cursor] > '9') {
      advance();
    }
    n = 0;
    while ('0' <= buffer[cursor] && buffer[cursor] <= '9') {
      n = n * 10 + buffer[cursor] - '0';
      advance();
    }
    return *this;
  }

 private:
  FILE *input_file;
  static const int SIZE = 1 << 17;
  int cursor;
  char buffer[SIZE];
  inline void advance() {
    ++cursor;
    if (cursor == SIZE) {
      cursor = 0;
      fread(buffer, SIZE, 1, input_file);
    }
  }
};
const int MAXVAL = 5000;
int v[1 + 3 * MAXVAL];
void binary(int x, int bits) {
  for (int i = 0; i < bits; i++)
    if (x & (1 << i))
      cout << "1";
    else
      cout << "0";
  cout << "    ";
}
void output(int a, int b, int c, int bits) {
  cout << a << "   " << b << "   " << c << "             ";
  binary(a, bits);
  binary(b, bits);
  binary(c, bits);
  cout << "\n";
}
long long get(long long n) {
  int r = n % 3;
  if (n == 1) return 1;
  if (n == 2) return 2;
  if (n == 3) return 3;
  int power = 1;
  while ((1LL << (2 * power)) - 1 < n) power++;
  n = (n - (1LL << (2 * (power - 1)))) / 3;
  long long a = 1LL << (2 * power - 2), b = 1LL << (2 * power - 1);
  for (int i = 0; i < power - 1; i++) {
    if (((n >> (2 * i)) & 3) == 0) {
    }
    if (((n >> (2 * i)) & 3) == 1) {
      a |= 1LL << (2 * i);
      b |= 1LL << (2 * i + 1);
    }
    if (((n >> (2 * i)) & 3) == 2) {
      a |= 1LL << (2 * i + 1);
      b |= 1LL << (2 * i);
      b |= 1LL << (2 * i + 1);
    }
    if (((n >> (2 * i)) & 3) == 3) {
      a |= 1LL << (2 * i);
      a |= 1LL << (2 * i + 1);
      b |= 1LL << (2 * i);
    }
  }
  if (r == 1) return a;
  if (r == 2) return b;
  if (r == 0) return a ^ b;
}
int main() {
  int tests;
  cin >> tests;
  for (int test = 1; test <= tests; test++) {
    long long n;
    cin >> n;
    cout << get(n) << "\n";
  }
  return 0;
}
