Problem: You are given an array a of length n consisting of zeros. You perform n actions with this array: during the i-th action, the following sequence of operations appears:

  1. Choose the maximum by length subarray (continuous subsegment) consisting only of zeros, among all such segments choose the leftmost one; 
  2. Let this segment be [l; r]. If r-l+1 is odd (not divisible by 2) then assign (set) a[(l+r)/(2)] := i (where i is the number of the current action), otherwise (if r-l+1 is even) assign (set) a[(l+r-1)/(2)] := i. 



Consider the array a of length 5 (initially a=[0, 0, 0, 0, 0]). Then it changes as follows:

  1. Firstly, we choose the segment [1; 5] and assign a[3] := 1, so a becomes [0, 0, 1, 0, 0]; 
  2. then we choose the segment [1; 2] and assign a[1] := 2, so a becomes [2, 0, 1, 0, 0]; 
  3. then we choose the segment [4; 5] and assign a[4] := 3, so a becomes [2, 0, 1, 3, 0]; 
  4. then we choose the segment [2; 2] and assign a[2] := 4, so a becomes [2, 4, 1, 3, 0]; 
  5. and at last we choose the segment [5; 5] and assign a[5] := 5, so a becomes [2, 4, 1, 3, 5]. 



Your task is to find the array a of length n after performing all n actions. Note that the answer exists and unique.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — the array a of length n after performing n actions described in the problem statement. Note that the answer exists and unique.

Example

Input


6
1
2
3
4
5
6


Output


1 
1 2 
2 1 3 
3 1 2 4 
2 4 1 3 5 
3 4 1 5 2 6 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
int noofdig(int N) { return floor(log10(N)) + 1; }
bool ispow2(int x) { return x && (!(x & (x - 1))); }
bool isVowel(char c) {
  if (c == 'A' || c == 'a' || c == 'E' || c == 'e' || c == 'I' || c == 'i' ||
      c == 'o' || c == 'O' || c == 'u' || c == 'U')
    return true;
  return false;
}
bool issqr(float y) {
  if (sqrt(abs(y)) == floor(sqrt(abs(y)))) return true;
  return false;
}
void printDivisors(int n) {
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        printf("%d ", i);
      else
        printf("%d %d ", i, n / i);
    }
  }
}
void removeSubstrs(string& s, string& p) {
  string::size_type n = p.length();
  for (string::size_type i = s.find(p); i != string::npos; i = s.find(p))
    s.erase(i, n);
}
int countone(unsigned int u) {
  unsigned int uCount;
  uCount = u - ((u >> 1) & 033333333333) - ((u >> 2) & 011111111111);
  return ((uCount + (uCount >> 3)) & 030707070707) % 63;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    vector<int> a(n, 0);
    priority_queue<pair<int, pair<int, int>>> q;
    q.push({n, {0, (n - 1)}});
    int cnt = 0;
    while (!q.empty()) {
      int l = -1 * q.top().second.first;
      int r = q.top().second.second;
      int len = q.top().first;
      q.pop();
      if (l > r) continue;
      cnt++;
      int m = (r + l) >> 1;
      a[m] = cnt;
      q.push({m - l + 1, {(-1 * l), m - 1}});
      q.push({r - m + 1, {-1 * (m + 1), r}});
    }
    for (auto& y : a) cout << y << " ";
    cout << "\n";
  }
  return 0;
}