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 Solution {
	
	
	
	static class subArr implements Comparable<subArr>{
		
		int f  ;
		int l;
		
		public int compareTo(subArr o)
		{
			
			int len2 =o.l-o.f+1;
			int len= this.l-this.f+1;
			
			if(len>len2)
			{
				return -1;
			}
			else if(len<len2) {
				return 1;
			}
			else {
				
				return this.f-o.f;
				
				
				
			}
			
		}
		
	}
	

	public static void main(String [] args)
	{
		Scanner in = new Scanner(System.in);
		
		int t =in.nextInt();
		
		for(int i =0;i<t;i++)
		{
			int n =in.nextInt();
			
			int [] a =new int [n];
			
			int count =1;
			
			
			PriorityQueue<subArr> myQueue = new PriorityQueue<>();
			
			subArr fArr =new subArr();
			fArr.f =0;
			fArr.l =n-1;
			myQueue.add(fArr);
			
			while(!myQueue.isEmpty())
			{
				subArr temp =myQueue.poll();
				subArr tempB =new subArr();
				subArr tempC =new subArr();
				int mid =(temp.f+temp.l)/2;
				a[mid] =count++;
				
				tempB.f =temp.f;
				tempB.l =mid-1;
				if(tempB.f<=tempB.l)
				myQueue.add(tempB);
				
				tempC.f =mid+1;
				tempC.l =temp.l;
				if(tempC.f<=tempC.l)
				myQueue.add(tempC);
				
				
			
			}
		
			for(int j =0;j<n;j++)
			{
			System.out.print(a[j]+" ");	
			}
		}
		
		
	}
		
		
	

}