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;
char buf[1 << 15], *fs, *ft;
inline char getc() {
  return (fs == ft &&
          (ft = (fs = buf) + fread(buf, 1, 1 << 15, stdin), fs == ft))
             ? 0
             : *fs++;
}
inline long long read() {
  register long long x = 0, f = 1;
  register char ch = getc();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getc();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getc();
  }
  return x * f;
}
const long long MAXN = 100010;
long long n, T, top;
long long s[MAXN], a, b, c;
long long p[MAXN], maxx, sum[MAXN];
int main() {
  T = read();
  p[0] = 1;
  maxx = 27;
  sum[0] = 1;
  for (register long long i = 1; i <= maxx; ++i)
    p[i] = p[i - 1] * 4, sum[i] += sum[i - 1] + p[i];
  while (T--) {
    n = read();
    top = 0;
    long long ww = (n - 1) / 3 + 1;
    long long cc = n % 3 == 0 ? 3 : n % 3;
    long long w1 = 0;
    while (sum[w1] < ww) ++w1;
    ww = ww - (w1 == 0 ? 0 : sum[w1 - 1]);
    while (w1) {
      long long p1 = (ww - 1) / 4 + 1;
      s[++top] = ww - (p1 - 1) * 4;
      ww = p1;
      --w1;
    }
    a = 1;
    b = 2;
    c = 3;
    while (top) {
      a = a << 2;
      b = b << 2;
      c = c << 2;
      if (s[top] == 2) a += 1, b += 2, c += 3;
      if (s[top] == 3) a += 2, b += 3, c += 1;
      if (s[top] == 4) a += 3, b += 1, c += 2;
      --top;
    }
    if (cc == 1) printf("%lld\n", a);
    if (cc == 2) printf("%lld\n", b);
    if (cc == 3) printf("%lld\n", c);
  }
  return 0;
}
