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.*;
public class test
{
public static void main(String[] args)
{
 Scanner sc=new Scanner(System.in);
 int t=sc.nextInt();
 for(int ii=0;ii<t;ii++)
 {
     int n=sc.nextInt();int ar[]=new int[n];
     PriorityQueue<pair> q=new PriorityQueue<pair>(new Comparator<pair>()
     {
         public int compare(pair one,pair two)
         {
             if(one.y-one.x>two.y-two.x)
             {
                 return -1;
             }
             else if(one.y-one.x<two.y-two.x)
             {
                 return 1;
             }
             else
             {
                 return one.x-two.x;
             }
         }}
         );
     q.add(new pair(1,n));int count=1;
     while(!q.isEmpty())
     {
         pair g=q.poll();
         int e=g.x;
         int f=g.y;int mi=0;
         if(e-f+1%2!=0)
          mi=(e+f)/2;
         else
         mi=(e+f-1)/2;
         ar[mi-1]=count++;
         if(count>n)
         break;
         
         if(mi+1>=1&&mi-1<=n&&f<=n&&f>=1)
         q.add(new pair(mi+1,f));
         if(mi-1>=1&&mi-1<=n&&e>=1&&e<=n)
         q.add(new pair(e,mi-1));
         
     }
     for(int i=0;i<ar.length;i++)
     System.out.print(ar[i]+" ");
     System.out.println();
 }
}}
class pair
{
    int x;
    int y;
    pair(int a,int b)
    {
        x=a;
        y=b;
    }
}