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 double TL = 1.5 * CLOCKS_PER_SEC;
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
vector<long long> get_row(long long r) {
  long long p = 1, s = 0, l = 0;
  for (;; p *= 4) {
    if (s + p >= r) {
      break;
    }
    l += 2;
    s += p;
  }
  long long d = r - s - 1;
  long long a = d + p;
  long long b = 2LL << l;
  for (int i = 0; i < l; i += 2) {
    long long x = (d % (1LL << (i + 2))) / (1LL << i);
    long long y = (d % (1LL << (i + 2))) / (1LL << i);
    if (x == 2 || x == 3) {
      b |= (1LL << i);
    }
    if (y == 1 || y == 2) {
      b |= (1LL << (i + 1));
    }
  }
  long long c = a ^ b;
  return {a, b, c};
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    long long n;
    cin >> n;
    n -= 1;
    long long x = n / 3 + 1, y = n % 3;
    vector<long long> ans = get_row(x);
    cout << ans[y] << endl;
  }
  return 0;
}
