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>
using namespace std;
class comp {
 public:
  bool operator()(pair<int, int> a, pair<int, int> b) {
    if ((a.second - a.first) < (b.second - b.first))
      return true;
    else if ((a.second - a.first) == (b.second - b.first))
      return a.first > b.first;
    else
      return false;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T, n;
  priority_queue<pair<int, int>, vector<pair<int, int>>, comp> q;
  cin >> T;
  while (T--) {
    cin >> n;
    int a[n];
    q.push({0, n - 1});
    int i = 1;
    while (!q.empty()) {
      if ((q.top().second - q.top().first) % 2 == 0) {
        a[(q.top().first + q.top().second) / 2] = i;
        if (q.top().first <= (q.top().first + q.top().second) / 2 - 1)
          q.push(make_pair(q.top().first,
                           (q.top().first + q.top().second) / 2 - 1));
        if ((q.top().first + q.top().second) / 2 + 1 <= q.top().second)
          q.push(make_pair((q.top().first + q.top().second) / 2 + 1,
                           q.top().second));
        q.pop();
        i++;
      } else {
        a[(q.top().second + q.top().first - 1) / 2] = i;
        if (q.top().first <= (q.top().first + q.top().second - 1) / 2 - 1)
          q.push(make_pair(q.top().first,
                           (q.top().first + q.top().second - 1) / 2 - 1));
        if ((q.top().first + q.top().second - 1) / 2 + 1 <= q.top().second)
          q.push(make_pair((q.top().first + q.top().second - 1) / 2 + 1,
                           q.top().second));
        q.pop();
        i++;
      }
    }
    for (int i = 0; i < n; i++) {
      cout << a[i] << ' ';
    }
    cout << endl;
  }
  return 0;
}