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 JAVA):

Solution:
import java.util.*;
import java.io.*;

public class D {
	


   public static void main(String[] args) throws IOException {
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
      int T = Integer.parseInt(br.readLine());
      for (int cases = 1; cases <= T; cases++) {
         int N = Integer.parseInt(br.readLine());
         int I = 1;
         int[] data = new int[N];
         PriorityQueue<Helper> pq = new PriorityQueue<Helper>();
         pq.add(new Helper(0, N-1));
         while (!pq.isEmpty()) {
            Helper h = pq.poll();
            if (h.left == h.right) {
               data[h.left] = I;
               I++;
            }
            else if (h.left < h.right) {
               if ((h.right - h.left + 1) % 2 == 1) {
                  int mid = (h.left + h.right)/2;
                  data[mid] = I;
                  I++;
                  pq.add(new Helper(h.left, mid-1));
                  pq.add(new Helper(mid+1, h.right));
               } 
               else {
                  int mid = (h.left + h.right-1)/2;
                  data[mid] = I;
                  I++;
                  pq.add(new Helper(h.left, mid-1));
                  pq.add(new Helper(mid+1, h.right));               
               }
            }
         }
         for (int i : data) {
            pw.print(i + " ");
         }
         pw.println();
      }
      pw.close();
   }
	
   static class Helper implements Comparable<Helper> {
   
      int left;
      int right;
      
      public Helper(int a, int b) {
         left = a;
         right = b;
      }
      
      public int compareTo(Helper h) {
         if (length() != h.length())
            return h.length() - length();
         return left - h.left;
      }
      
      private int length() {
         return (right-left+1);
      }
   
   }
	
	
}