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.*;
import java.math.*;

   public class D
   {
      public static void main(String omkar[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int T = Integer.parseInt(st.nextToken());
         StringBuilder sb = new StringBuilder();
         while(T-->0)
         {
            int N = Integer.parseInt(infile.readLine());
            int[] res = new int[N];
            PriorityQueue<Range> pq = new PriorityQueue<Range>();
            pq.add(new Range(0, N-1));
            int tag = 1;
            while(pq.size() > 0)
            {
               Range curr = pq.poll();
               int dist = curr.right-curr.left+1;
               if(dist == 1)
                  res[curr.left] = tag;
               else
               {
                  if(dist%2 == 1)
                  {
                     res[dist/2+curr.left] = tag;
                     if(curr.left <= dist/2+curr.left-1)
                        pq.add(new Range(curr.left, dist/2+curr.left-1));
                     if(dist/2+curr.left+1 <= curr.right)
                        pq.add(new Range(dist/2+curr.left+1, curr.right));
                  }
                  else
                  {
                     res[dist/2+curr.left-1] = tag;
                     if(curr.left <= dist/2+curr.left-2)
                        pq.add(new Range(curr.left, dist/2+curr.left-2));
                     if(dist/2+curr.left <= curr.right)
                        pq.add(new Range(dist/2+curr.left, curr.right));
                  }
               }
               tag++;
            }
            for(int x: res)
               sb.append(x+" ");
            sb.append("\n");
         }
         System.out.print(sb);
      }
   }
   class Range implements Comparable<Range>
   {
      public int left;
      public int right;
      
      public Range(int a, int b)
      {
         left = a;
         right = b;
      }
      public int compareTo(Range oth)
      {
         if(right-left == oth.right-oth.left)
            return left-oth.left;
         return (oth.right-oth.left)-(right-left);
      }
   }