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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  for (int z = 0; z < t; ++z) {
    long long n;
    cin >> n;
    long long k = 0;
    long long cur = 0;
    long long m = (n + 2) / 3;
    while (cur + (1ll << (2 * k)) < m) {
      cur += (1ll << (2 * k));
      k++;
    }
    m -= cur;
    vector<int> b;
    m--;
    while (m > 0) {
      b.push_back(m % 2);
      m /= 2;
    }
    while (b.size() < 2 * k) b.push_back(0);
    reverse(b.begin(), b.end());
    vector<int> b1(2 * k), b2(2 * k);
    for (int i = 0; i < k; ++i) {
      if (b[i * 2] == 0 && b[i * 2 + 1] == 0) {
        b1[i * 2] = 0;
        b1[i * 2 + 1] = 0;
        b2[i * 2] = 0;
        b2[i * 2 + 1] = 0;
      } else if (b[i * 2] == 0 && b[i * 2 + 1] == 1) {
        b1[i * 2] = 1;
        b1[i * 2 + 1] = 0;
        b2[i * 2] = 1;
        b2[i * 2 + 1] = 1;
      } else if (b[i * 2] == 1 && b[i * 2 + 1] == 0) {
        b1[i * 2] = 1;
        b1[i * 2 + 1] = 1;
        b2[i * 2] = 0;
        b2[i * 2 + 1] = 1;
      } else if (b[i * 2] == 1 && b[i * 2 + 1] == 1) {
        b1[i * 2] = 0;
        b1[i * 2 + 1] = 1;
        b2[i * 2] = 1;
        b2[i * 2 + 1] = 0;
      }
    }
    long long ans;
    if (n % 3 == 1) {
      ans = 1;
      for (int i = 0; i < 2 * k; ++i) {
        ans *= 2;
        ans += b[i];
      }
    } else if (n % 3 == 2) {
      ans = 2;
      for (int i = 0; i < 2 * k; ++i) {
        ans *= 2;
        ans += b1[i];
      }
    } else if (n % 3 == 0) {
      ans = 3;
      for (int i = 0; i < 2 * k; ++i) {
        ans *= 2;
        ans += b2[i];
      }
    }
    cout << ans << "\n";
  }
  return 0;
}
