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 TP1 {
	static class pair{
		int l,r,size;
		public pair(int l,int r) {
			this.l=l;
			this.r=r;
			this.size=r-l+1;
		}
	}
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0)
		{
			int n=sc.nextInt();
			int a[] = new int[n+1];
			PriorityQueue<pair> p=new PriorityQueue<>(new Comparator<pair>() {			
				public int compare(pair o1, pair o2) 
				{
					if(o1.size==o2.size)
					{
						return o1.l-o2.l;
					}
					else
					{
						return o2.size-o1.size;
					}
				}
			});
			p.add(new pair(1, n));
			int k=1;
			for(int i=1;i<=n;i++)
			{
				pair p1=p.poll();
				int m=p1.l + ((p1.r-p1.l)/2);
				a[m]=k;
				k++;
				p.add(new pair(p1.l, m-1));
				p.add(new pair(m+1,p1.r));
			}
			for(int i=1;i<=n;i++)
			{
				System.out.print(a[i]+" ");
			}
		}
	}
}