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 codeforces
{
	public static void main(String[] args) 
	{
		Scanner s=new Scanner(System.in);
		int t=s.nextInt();

		for(int k=0;k<t;k++)
		{
			int n=s.nextInt();
			int start=1,end=n;

			int temp[]={1,n,n};
			PriorityQueue<int[]>pq=new PriorityQueue<>((a,b)->a[2]==b[2]?a[0]-b[0]:b[2]-a[2]);

			int outArr[]=new int[n+1];
			pq.add(temp);
			int count =1;
			while(pq.size()>0)
			{
				int arr[]=pq.remove();
				start=arr[0];
				end=arr[1];

				int  mid=0;
				if(start==end)
				{
					outArr[start]=count++;
				}
				else
				{
					mid=(start+end)/2;
					outArr[mid]=count++;
					if(mid-start>0)
						pq.add(new int[]{start,mid-1,mid-start});
					if(end-mid>0)
						pq.add(new int[]{mid+1,end,end-mid});
				}
			}
			for(int i=1;i<=n;i++)
				System.out.print(outArr[i]+" ");
		}	
	}
}