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.Comparator;
import java.util.PriorityQueue;
import java.util.Scanner;
public class D642 
{
	public static void main(String[] args)
	{
		Scanner obj=new Scanner(System.in);
		int t=obj.nextInt();
		while(t-->0)
		{
			int n=obj.nextInt();
		PriorityQueue<pair> q=new PriorityQueue<>(new comparator());
			q.add(new pair(1,n));
			int arr[]=new int[n+1];
			int i=1;
			while(!q.isEmpty())
			{
				pair p=q.poll();
				int mid=(p.x+p.y)/2;
				arr[mid]=i;
				i++;
				if(mid>p.x)
					q.add(new pair(p.x,mid-1));
				if(mid<p.y)
					q.add(new pair(mid+1,p.y));
			}
			for(int j:arr)
			{
				if(j!=0)
					System.out.print(j+" ");
			}
			System.out.println();
		}
	}
}
class pair {
	int x;
	int y;
	public pair(int x,int y)
	{
		this.x=x;
		this.y=y;

	}


}

class comparator implements Comparator<pair>
{
	@Override
		public int compare(pair a, pair b)
		{
			int lena=a.y-a.x+1;
			int lenb=b.y-b.x+1;
			if(lena==lenb)
			{
				 return a.x-b.x;
			}
			return lenb-lena;
		}
}