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;
const int MOD = 1e9 + 7;
void build(int l, int r, vector<pair<int, int> >& v) {
  if (l > r) return;
  int p = (l + r) / 2;
  v.push_back({(r - l + 1), p});
  build(l, p - 1, v);
  build(p + 1, r, v);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  ;
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    vector<pair<int, int> > v;
    build(0, n - 1, v);
    assert(v.size() == n);
    sort(v.begin(), v.end(),
         [](const pair<int, int>& x, const pair<int, int>& y) {
           if (x.first == y.first) return x.second < y.second;
           return x.first > y.first;
         });
    vector<long long> ans(n);
    for (int i = 0; i < v.size(); i++) {
      ans[v[i].second] = i + 1;
    }
    for (auto i : ans) cout << i << " ";
    cout << '\n';
  }
  return 0;
}