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;
int t;
long long n, sum, x, multi, ans;
vector<pair<int, int> > v;
int main() {
  int i, k;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld", &n);
    sum = 0;
    for (i = 0;; i++) {
      sum += 3 * (1LL << (i * 2));
      if (sum >= n) break;
    }
    x = 3 * (1LL << (i * 2));
    sum -= x;
    n -= sum;
    v.clear();
    if (n % 3 == 1)
      v.emplace_back(0, 1);
    else if (n % 3 == 2)
      v.emplace_back(1, 0);
    else
      v.emplace_back(1, 1);
    k = i;
    for (i = 1; i <= k; i++) {
      if (n <= x / 4)
        v.emplace_back(0, 0);
      else if (n <= x / 4 * 2) {
        if (n % 3 == 1)
          v.emplace_back(0, 1);
        else if (n % 3 == 2)
          v.emplace_back(1, 0);
        else
          v.emplace_back(1, 1);
        n -= x / 4;
      } else if (n <= x / 4 * 3) {
        if (n % 3 == 1)
          v.emplace_back(1, 0);
        else if (n % 3 == 2)
          v.emplace_back(1, 1);
        else
          v.emplace_back(0, 1);
        n -= x / 2;
      } else {
        if (n % 3 == 1)
          v.emplace_back(1, 1);
        else if (n % 3 == 2)
          v.emplace_back(0, 1);
        else
          v.emplace_back(1, 0);
        n -= x / 4 * 3;
      }
      x /= 4;
    }
    multi = 1;
    ans = 0;
    for (i = v.size() - 1; i >= 0; i--) {
      ans += multi * (2 * v[i].first + v[i].second);
      multi *= 4;
    }
    printf("%lld\n", ans);
  }
  return 0;
}
