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.io.*;
import java.util.*;

public class D {

	public static void main(String[] args)throws IOException {
		
		BufferedReader ob=new BufferedReader(new InputStreamReader(System.in));
		StringBuffer sb=new StringBuffer();
		
		int test=Integer.parseInt(ob.readLine());
		while(test-->0)
		{
			int n=Integer.parseInt(ob.readLine());
			
			int ans[] = new int[n];
			int index = 1;
			
			PriorityQueue<int[]> pq=new PriorityQueue<int[]>(new Comparator<int[]>()
			{
				public int compare(int[] a, int[] b) 
				{
					int as=a[1]-a[0];
					int bs=b[1]-b[0];
//					System.out.println("as , bs = "+as+" , "+bs);
					if (as!=bs)
						return (bs-as);
					else
						return a[0]-b[0];
				}
			});
			
			pq.add(new int[]{0,n-1});
			while (! pq.isEmpty()) 
			{
				int rng[] = pq.poll();
				int join = (rng[0]+rng[1])/2;
				ans[join] = index;
				index++;
				if (rng[1]-rng[0] > 1) 
				{
					pq.add(new int[]{rng[0],join-1});
					pq.add(new int[]{join+1,rng[1]});
				} 
				else if (rng[1]-rng[0] == 1) 
					pq.add(new int[]{join+1,rng[1]});
//				System.out.println("pq = "+pq);
			}
			
			for(int i=0;i<n;i++)
			{
				sb.append(ans[i]+" ");
			}
			sb.append("\n");
		}
		System.out.println(sb);
	}
}