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;
struct cmp {
  bool operator()(const pair<long long, long long> &a,
                  const pair<long long, long long> &b) const {
    return (a.second - a.first == b.second - b.first)
               ? (a.first < b.first)
               : (a.second - a.first > b.second - b.first);
  }
};
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    set<pair<long long, long long>, cmp> q((cmp()));
    vector<long long> ans(n, 0);
    long long last = 0;
    q.insert({0, n});
    while (!q.empty()) {
      auto [l, r] = *q.begin();
      q.erase(q.begin());
      long long m = (l + r - 1) / 2;
      ans[m] = ++last;
      long long nl1 = l;
      long long nr1 = m;
      long long nl2 = m + 1;
      long long nr2 = r;
      if (nr1 > nl1) q.insert({nl1, nr1});
      if (nr2 > nl2) q.insert({nl2, nr2});
    }
    for (long long e : ans) cout << e << " ";
    cout << "\n";
  }
}