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.*;

public class Main
{
	public static void main(String[] args) {
	    Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		while (t-- > 0) {
		    int n = in.nextInt();
		    int[] a = new int[n];
		    
		    a[(n-1)/2] = 1;
		    
		    PriorityQueue<Bounds> queue = new PriorityQueue<>();
		    queue.add(new Bounds(0,(n-1)/2 - 1));
		    queue.add(new Bounds((n-1)/2 + 1, n-1));
		    
		    int next = 2;
		    while (next <= n) {
		        Bounds b = queue.poll();
		        a[b.midpt()] = next++;
		        
		        if (b.width > 0) {
		            queue.add(new Bounds(b.left, b.midpt()-1));
		            queue.add(new Bounds(b.midpt()+1, b.right));
		        }
		    }
		    
		    for (int e : a) {
		        System.out.print(e + " ");
		    }
		    System.out.println();
		}
	}
}

class Bounds implements Comparable<Bounds> {
    int left, right, width;
    public Bounds(int l, int r) {
        left = l; right = r; width = (r-l+1);
    }
    public int midpt() {
        return (right+left)/2;
    }
    public int compareTo(Bounds o) {
        if (width == o.width)
            return left - o.left;
        return o.width - width;
    }
}