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 solve() {
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) {
    a[i] = 0;
  }
  long long count = 1;
  int middle;
  set<pair<int, pair<int, int>>> s{};
  s.insert(make_pair(1 - n, make_pair(0, n - 1)));
  while (s.size() != 0) {
    int r = (*(s.begin())).second.second;
    int l = (*(s.begin())).second.first;
    s.erase(s.begin());
    if ((r - l + 1) % 2 == 0) {
      middle = (r + l - 1) / 2;
    } else {
      middle = (l + r) / 2;
    }
    if (r >= l) {
      a[middle] = count++;
      s.insert(make_pair(-1 * (r - middle - 1), make_pair(middle + 1, r)));
      s.insert(make_pair(-1 * (middle - 1 - l), make_pair(l, middle - 1)));
    }
  }
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  }
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}