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;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    if (n == 1) {
      cout << 1 << "\n";
      continue;
    }
    int moves = 1;
    priority_queue<pair<int, pair<int, int> > > pq;
    int arr[n + 1];
    if (n % 2) {
      arr[n / 2 + 1] = 1;
      long long news = n / 2;
      pq.push(make_pair(n / 2, make_pair(1, n / 2)));
      pq.push(make_pair(n / 2, make_pair(n / 2 + 2, n)));
    } else {
      arr[n / 2] = 1;
      pq.push(make_pair(n / 2, make_pair(n / 2 + 1, n)));
      pq.push(make_pair(n / 2 - 1, make_pair(1, n / 2 - 1)));
    }
    moves = 1;
    while (!pq.empty()) {
      int kk = pq.top().first;
      int lll = pq.top().second.first;
      int rr = pq.top().second.second;
      if (kk == 0) break;
      vector<pair<int, pair<int, int> > > v;
      while (kk == pq.top().first && !pq.empty()) {
        pair<int, pair<int, int> > pp;
        pp = pq.top();
        pq.pop();
        v.push_back(pp);
      }
      sort(v.begin(), v.end());
      for (int i = 0; i < v.size(); i++) {
        moves++;
        pair<int, pair<int, int> > ss;
        ss = v[i];
        kk = v[i].first;
        lll = v[i].second.first;
        rr = v[i].second.second;
        if (kk == 1) {
          arr[lll] = moves;
          continue;
        }
        if (kk == 0) break;
        if (kk % 2) {
          arr[lll + kk / 2] = moves;
          long long news = kk / 2;
          pq.push(make_pair(kk / 2, make_pair(lll, lll + kk / 2 - 1)));
          pq.push(make_pair(kk / 2, make_pair(lll + kk / 2 + 1, rr)));
        } else {
          arr[lll + kk / 2 - 1] = moves;
          pq.push(make_pair(kk / 2, make_pair(lll + kk / 2, rr)));
          pq.push(make_pair(kk / 2 - 1, make_pair(lll, lll + kk / 2 - 1)));
        }
      }
    }
    for (int i = 1; i <= n; i++) cout << arr[i] << " ";
    cout << "\n";
  }
}