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() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int arr[n + 1];
    queue<pair<int, int>> q;
    q.push({1, n});
    int ct = 1;
    while (!q.empty()) {
      priority_queue<int, vector<int>, greater<int>> pq;
      pair<int, int> temp2 = q.front();
      q.pop();
      int len = temp2.second - temp2.first + 1;
      pq.push(temp2.first);
      queue<pair<int, int>> q2;
      while (!q.empty()) {
        pair<int, int> temp1 = q.front();
        int len1 = temp1.second - temp1.first + 1;
        if (len1 != len) {
          q2.push(q.front());
          q.pop();
        } else {
          pq.push(temp1.first);
          q.pop();
        }
      }
      while (!q2.empty()) {
        q.push(q2.front());
        q2.pop();
      }
      while (!pq.empty()) {
        int idx;
        int x = pq.top();
        pq.pop();
        pair<int, int> temp;
        temp.first = x;
        temp.second = x + len - 1;
        if (len % 2 == 0) {
          idx = (temp.second + temp.first - 1) / 2;
          arr[idx] = ct;
          ct++;
        } else {
          idx = (temp.second + temp.first) / 2;
          arr[idx] = ct;
          ct++;
        }
        if (len == 2) {
          q.push({idx + 1, temp.second});
        } else if (len != 1 && len % 2 == 0) {
          q.push({idx + 1, temp.second});
          q.push({temp.first, idx - 1});
        } else if (len != 1 && len % 2 != 0) {
          q.push({temp.first, idx - 1});
          q.push({idx + 1, temp.second});
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      cout << arr[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}