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.LinkedList;

import java.util.*;
public class shortpath {
	static class pair{
		int l,r,size;
		
		pair(int l,int r){
			this.l=l;
			this.r=r;
			this.size=r-l+1;
		}
		
		@Override
		public String toString() {
			return l+" "+r+" "+size;
		}
		
	}
public static void main(String[] args) {
	Scanner scan=new Scanner(System.in);
	
	int t =scan.nextInt();
while(t>0) {	
	int n=scan.nextInt();
	PriorityQueue<pair>p=new PriorityQueue<shortpath.pair>(new Comparator<pair>() {

		@Override
		public int compare(pair o1, pair o2) {
		if(o1.size==o2.size) {
			return o1.l-o2.l;
		}
		return o2.size-o1.size;
		
		}
	});

	p.add(new pair(1,n));
	int [] a=new int[n+1];
	int i;
for(i=1;i<=n;i++) {
	pair p1=p.poll();
	
	int l=p1.l;
	int r=p1.r;
	int m;
	int result=r-l+1;
	
	if(result%2!=0) {
		m=(r+l)/2;
	}
	else
	m=(r+l-1)/2;
	
a[m]=i;

p.add(new pair(l,m-1));
p.add(new pair(m+1,r));
	
	
}

for(i=1;i<=n;i++) {
System.out.print(a[i]+" ");	
}

System.out.println();

t--;
}


}
}