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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) throws IOException{
		StreamTokenizer sc=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		PrintWriter out=new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		
		while (sc.nextToken()!=StreamTokenizer.TT_EOF) {
			int T=(int)sc.nval;
			while (T-->0) {
				sc.nextToken();
				int n=(int)sc.nval;
				int[] all=new int[n+1];
				int now=1;
				PriorityQueue<edge> temp=new PriorityQueue<edge>();
				temp.add(new edge(1, n));
				while (!temp.isEmpty()) {
					edge aaa=temp.poll();
					int r=aaa.r,l=aaa.l;
//					System.out.println(l+" "+r+" "+now);
					if (r>=all.length||r<=0||l<=0||l>=all.length) {
						continue;
					}
					if (now>=all.length) {
						break;
					}
					if (l==r) {
						all[l]=now;
					}
					int sum=l+r;
					if (sum%2==0) {
						int after=sum/2;
						all[after]=now;
						if (after-1>=l) {
							temp.add(new edge(l, after-1));
						}
						if (after+1<=r) {
							temp.add(new edge(after+1, r));
						}
						
					}else {
						int after=sum/2;
						all[after]=now;
						if (after+1<=r) {
							temp.add(new edge(after+1, r));
						}
						if (after-1>=l) {
							temp.add(new edge(l, after-1));
						}
						
					}
					now++;
					
				}
				for (int i = 1; i < all.length; i++) {
					if (i!=1) {
						out.print(' ');
					}
					out.print(all[i]);
				}
				out.println();
				out.flush();
				
			}

		}
		
	}

}
class edge implements Comparable<edge>{
	int l,r,len;
	public edge(int l,int r) {
		this.l=l;
		this.r=r;
		len=r-l;
	}
	@Override
	public int compareTo(edge o) {
		if (o.len==this.len) {
			return this.l-o.l;
		}
		return o.len-this.len;
	}
	
}