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;
void showpq(
    priority_queue<pair<long long int, pair<long long int, long long int> > >
        q) {
  priority_queue<pair<long long int, pair<long long int, long long int> > > qq =
      q;
  while (!qq.empty()) {
    cout << qq.top().first << " " << qq.top().second.first << " "
         << qq.top().second.second << endl;
    qq.pop();
  }
  cout << '\n';
}
int main() {
  long long int T;
  scanf("%I64d", &T);
  while (T--) {
    long long int n, k, m, t, i, j, res = 0, mid, t2, dist;
    scanf("%I64d", &n);
    long long int arr[n + 5];
    for (i = 0; i < n + 2; i++) arr[i] = 0;
    queue<pair<long long int, long long int> > qq;
    priority_queue<pair<long long int, pair<long long int, long long int> > >
        pq;
    t = 1;
    qq.push(make_pair(1, n));
    t2 = 0;
    while (!qq.empty()) {
      pair<long long int, long long int> pp, p2, p3;
      pp = qq.front();
      qq.pop();
      dist = pp.second - pp.first;
      if (pp.first == pp.second) {
        arr[pp.first] = t;
      } else if ((pp.first + pp.second) % 2 == 0) {
        mid = (pp.first + pp.second) / 2;
        arr[mid] = t;
        if ((mid + 1) <= pp.second)
          pq.push(
              make_pair(pp.second - mid - 1, make_pair(-mid - 1, pp.second)));
        if (pp.first <= mid - 1)
          pq.push(make_pair(mid - 1 - pp.first, make_pair(-pp.first, mid - 1)));
      } else {
        mid = (pp.first + pp.second - 1) / 2;
        arr[mid] = t;
        if ((mid + 1) <= pp.second)
          pq.push(
              make_pair(pp.second - mid - 1, make_pair(-mid - 1, pp.second)));
        if (pp.first <= mid - 1)
          pq.push(make_pair(mid - 1 - pp.first, make_pair(-pp.first, mid - 1)));
      }
      if (!pq.empty()) {
        p3 = pq.top().second;
        p3.first *= (-1);
        qq.push(p3);
        pq.pop();
      }
      t++;
    }
    for (i = 1; i <= n; i++) cout << arr[i] << " ";
    cout << endl;
  }
}