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 D3D {
	public static void main(String[] args) throws Exception{
		int num = 998244353;
		// TODO Auto-generated method stub
 		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
 		PrintWriter out = new PrintWriter(System.out);
 		int t = Integer.parseInt(bf.readLine());
 		for(int i = 0;i<t;i++){
 			
 			int count = 0;
 			int a = Integer.parseInt(bf.readLine());
 			int[] array = new int[a];
 			PriorityQueue<Coord> q = new PriorityQueue<Coord>();
 			q.add(new Coord(0, a-1));
 			while (!q.isEmpty()){
 				Coord c = q.remove();
 				count+=1;
 				int l = c.getL();
 				int r = c.getR();
 				if (r >= l ){
 					array[(l+r)/2] = count;
 					if ((r-l)%2 == 0){
 						q.add(new Coord(l, (r+l)/2-1));
 						q.add(new Coord((r+l)/2+1, r));
 					}
 					else{
 						q.add(new Coord((r+l)/2+1, r));
 						q.add(new Coord(l, (r+l)/2-1));
 					}
 				}
 					
 			}
 			for(int j = 0;j<a;j++){
 				out.print(array[j] + " ");
 			}
 			out.println();
 		}
 		out.close();
 			
 	}
 		
}
class Coord implements Comparable<Coord>{
	private int l;
	private int r;
	public Coord( int l, int r){
		this.l = l;
		this.r = r;
	}
	public int getL() {
		return l;
	}
	public void setL(int l) {
		this.l = l;
	}
	public int getR() {
		return r;
	}
	public void setR(int r) {
		this.r = r;
	}
	public int compareTo(Coord other){
		if (r-l > other.r - other.l)
			return -1;
		else if (r-l < other.r - other.l)
			return 1;
		else{
			if (l < other.l)
				return -1;
			else if (l > other.l)
				return 1;
			else
				return 0;
		}
	}
	
}