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() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> res(n, 0);
    map<long long, set<long long>> mp;
    mp[n].insert(0);
    long long value = 1, index_of, k, l, r, length, length1, length2;
    while (value <= n) {
      auto it = mp.rbegin();
      length = it->first;
      auto tmp1 = it->second.begin();
      l = *tmp1;
      r = l + length - 1;
      it->second.erase(it->second.begin());
      if (it->second.size() == 0) mp.erase(length);
      if (l == r)
        res[l] = value;
      else {
        if (r - l == 1) {
          res[l] = value;
          mp[1].insert(l + 1);
        } else {
          k = r - l + 1;
          if (k % 2 != 0)
            index_of = (l + r) / 2;
          else
            index_of = (l + r - 1) / 2;
          res[index_of] = value;
          length1 = index_of - l;
          length2 = r - index_of;
          mp[length1].insert(l);
          mp[length2].insert(index_of + 1);
        }
      }
      value++;
    }
    for (long long i = 0; i < n; i++) cout << res[i] << " ";
    cout << endl;
  }
}