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.*;
import java.io.*;
public class GFG {
	public static void main (String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while (t-->0)
		{
		    int n=sc.nextInt();
		    int[] arr=new int[n+1];
		    PriorityQueue<pair> pq=new PriorityQueue<pair>();
		    pq.add(new pair(1,n));
		    for(int i=1;i<=n;i++)
		    {
		        pair pr=pq.remove();
		        int mid=-1;
		        if(pr.left>pr.right)
		        continue;
		        if(pr.total%2==0)
		        mid=(pr.left+pr.right-1)/2;
		       if(pr.total%2!=0)
		        mid=(pr.left+pr.right)/2;
		        arr[mid]=i;
		        pq.add(new pair(pr.left,mid-1));
		        pq.add(new pair(mid+1,pr.right));
		    }
		    for(int i=1;i<=n;i++)
		    System.out.print(arr[i]+" ");
		    System.out.println();
		}
	}
	
	static class pair implements Comparable<pair>
	{
	    int left,right,total;
	    pair(int a,int b)
	    {left=a;right=b;total=right-left+1;}
	    @Override
	    public int compareTo(pair p)
	    {
	        if(this.total!=p.total)
	        return p.total-this.total;
	        return this.left-p.right;
	    }
	}
}