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.LinkedList;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main{

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scn=new Scanner(System.in);
		int t =scn.nextInt();
		while(t-->0){
			int n=scn.nextInt();
			int [] arr=new int[n];
			count=1;
		   res(n,arr);
		   StringBuilder sb=new StringBuilder();
		   for(int i=0;i<arr.length;i++){
			   sb.append(arr[i]+" ");
		   }
		   System.out.println(sb);
		}

	}
	static int count=0;
	
	public static void res(int n,int [] arr){
		PriorityQueue<pair> q =new PriorityQueue<>();
		
		q.add(new pair(0,n-1,n));
		int count=1;
		while(q.size()>0){
			if(count>n){
				break;
			}
			pair p = q.remove();
			if(p.s==p.e){
				arr[p.s]=count;
				count++;
				continue;
			}
		     int mid= (p.s+p.e)/2;
		    
		     if(arr[mid]!=0){
		    	 continue;
		     }
		     arr[mid]=count;
		     count++;
		     
		    	 q.add(new pair(p.s,mid-1,mid-p.s));
		    	 q.add(new pair(mid+1,p.e,p.e-mid));
		    
		}
		
	}
	public static class pair implements Comparable<pair>{
		int s;
		int e;
		int length;
		public  pair(int s ,int e,int length){
			this.s=s;
			this.e=e;
			this.length=length;
		}
		public int compareTo(pair o){
			if(this.length!=o.length){
				return o.length-this.length;
			}else{
				
					return this.s-o.s;
				
			}
		}
	}

}