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 int maxN = 1010;
int used[maxN];
void brut() {
  vector<int> res;
  int idx = 0;
  while (idx < 100) {
    ++idx;
    int found = false;
    for (int i = 1; i < maxN && !found; ++i) {
      for (int j = i; j < maxN && !found; ++j) {
        for (int k = j; k < maxN && !found; ++k) {
          if (!used[i] && !used[j] && !used[k]) {
            if ((i ^ j ^ k) == 0) {
              res.push_back(i);
              res.push_back(j);
              res.push_back(k);
              used[i] = used[j] = used[k] = 1;
              cout << i << " " << j << " " << k << endl;
              found = true;
            }
          }
        }
      }
    }
  }
}
long long f[16] = {0, 1, 2, 3, 4, 8, 12, 5, 10, 15, 6, 11, 13, 7, 9, 14};
vector<long long> genrow(long long bucket, long long row) {
  if (bucket == 1) {
    return {1, 2, 3};
  }
  long long sz = bucket / 4;
  auto res = genrow(bucket / 4, row % sz);
  long long order[4] = {0, 2, 3, 1};
  long long a = bucket + row;
  long long b = 2 * bucket + (res[1] - 2 * sz) + order[row / sz] * sz;
  return {a, b, a ^ b};
}
vector<long long> genrow(long long row) {
  if (row == 1) {
    return {1, 2, 3};
  }
  long long total = 1;
  long long cur_size = 1;
  while (total < row) {
    cur_size *= 4;
    total += cur_size;
  }
  long long pre_size = total - cur_size;
  return genrow(cur_size, row - pre_size - 1);
}
void solve() {
  long long n;
  cin >> n;
  long long row = (n + 2) / 3;
  auto res = genrow(row);
  if (n % 3 == 1) {
    cout << res[0] << endl;
  } else if (n % 3 == 2) {
    cout << res[1] << endl;
  } else {
    cout << res[2] << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  for (int tt = 0; tt < t; ++tt) {
    solve();
  }
  return 0;
}
