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;
vector<int> vtest;
void eprintb(int x) {
  for (int i = 20; i >= 0; i--)
    if (x >> i)
      printf("%d", (x >> i) & 1);
    else
      printf(" ");
  printf("\n");
}
const int N = (int)1e6 + 100;
bool used[N];
long long zzz[4] = {0, 2, 3, 1};
long long getX(int len, long long pos) {
  long long ans = (1LL << len);
  for (int i = 0; i < len; i += 2) {
    ans |= zzz[(pos >> i) & 3LL] << i;
  }
  return ans;
}
long long solve1(long long pos) {
  int len = 1;
  while (pos >= 3LL * (1LL << (len - 1))) {
    pos -= 3LL * (1LL << (len - 1));
    len += 2;
  }
  int rem = pos % 3;
  pos /= 3;
  long long val = (1LL << (len - 1)) + pos;
  if (rem == 0) return val;
  long long x = getX(len, pos);
  if (rem == 1) return x;
  if (rem == 2) return x ^ val;
  throw;
}
void solve() {
  long long pos;
  scanf("%lld", &pos);
  pos--;
  long long ans = solve1(pos);
  if (ans >= 0 && pos < (int)vtest.size() && vtest[pos] != ans) {
    42;
    eprintb(pos);
    eprintb(ans);
    eprintb(vtest[pos]);
    throw;
  }
  printf("%lld\n", ans);
}
int main(int, char **) {
  for (int i = 1; i < (int)100; i++) {
    if (used[i]) continue;
    used[i] = true;
    for (int j = i + 1; j < N; j++) {
      int h = i ^ j;
      if (used[j] || used[h]) continue;
      used[i] = used[j] = used[h] = true;
      vtest.push_back(i);
      vtest.push_back(j);
      vtest.push_back(h);
      break;
    }
  }
  int t;
  scanf("%d", &t);
  for (int i = 0; i < t; i++) solve();
  return 0;
}
