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.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.StringTokenizer; 
import java.util.*;
 
public class Solution 
{ 
    public static int[] ans(int n){
        int[] arr=new int[n];
         PriorityQueue<Segment> pQueue =  
             new PriorityQueue<Segment>(new SortSeg()); 
             Segment first=new Segment(0,n-1);
             pQueue.add(first);
             int num=1;
             while(pQueue.isEmpty()==false){
                 Segment p =pQueue.poll(); 
                 int s=p.s;
                 int e=p.e;
                 int m=(s+e)/2;
                 arr[m]=num;
                 num++;
                   Segment k1,k2;
                 if(s<m){
               k1=new Segment(s,m-1);
                   pQueue.add(k1);
             }
             if(m<e){
                   k2=new Segment(m+1,e);
                  pQueue.add(k2);
             }
             }
            return arr; 
    }
    public static void main(String[] args) throws IOException 
    { 
  
       BufferedReader br = new BufferedReader( 
                              new InputStreamReader(System.in));
        StringBuilder out=new StringBuilder(); 
        int t=Integer.parseInt(br.readLine());
      for(int j=0;j<t;j++){
            int n=Integer.parseInt(br.readLine());
            int a[]=ans(n);
            for(int i=0;i<n;i++){
                out.append(a[i]+" ");
            }
            out.append("\n");
        }
        
        
      
      System.out.println(out);
    }
}
 class Segment{
    int s;
    int e;
    public Segment(int x,int y){
        s=x;
        e=y;
    }
     
 }
class SortSeg implements Comparator<Segment> 
{ 
    // Used for sorting in ascending order of 
    // roll number 
    public int compare(Segment a, Segment b) 
    { 
        int lena= a.e-a.s+1;
         int lenb= b.e-b.s+1;
        if(lena==lenb)
        return a.s - b.s; 
        else
        //decsending order
        return lenb-lena;
    } 
}