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 option {
  bool operator()(const pair<int, pair<int, int>>& a,
                  const pair<int, pair<int, int>>& b) const {
    return a.first < b.first ||
           a.first == b.first && a.second.first > b.second.first;
  }
};
int main() {
  int T;
  cin >> T;
  while (T) {
    T--;
    int n;
    cin >> n;
    vector<int> ans(n);
    priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>,
                   option>
        a;
    a.push({n, {0, n - 1}});
    int i = 1;
    while (a.size() > 0) {
      int q = a.top().second.first;
      int p = a.top().second.second;
      int temp;
      a.pop();
      if (p < q || p < 0) continue;
      if ((p + q) % 2 == 0) {
        temp = (p + q) / 2;
        ans[temp] = i;
      } else {
        temp = (p + q - 1) / 2;
        ans[temp] = i;
      }
      a.push({temp - q, {q, temp - 1}});
      a.push({p - temp, {temp + 1, p}});
      i++;
    }
    for (int i = 0; i < ans.size(); i++) {
      cout << ans[i] << " ";
    }
    cout << endl;
  }
}