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 D {

	public static void main(String[] args) {
		Scanner file = new Scanner(System.in);
		int inputs = file.nextInt();
		while(inputs-->0) {
			int n = file.nextInt();
			int[] arr = new int[n];
			PriorityQueue<Segment> toDo = new PriorityQueue<>(Comparator.reverseOrder());
			toDo.add(new Segment(0, n-1));
			int num = 1;
			while(!toDo.isEmpty()) {
				Segment s = toDo.poll();
				if(s.length < 0)
					continue;
				arr[s.mid] = num++;
				if(s.length >= 1) {
					toDo.add(new Segment(s.left, s.mid-1));
					toDo.add(new Segment(s.mid+1, s.right));
				}
			}
			StringBuilder ans = new StringBuilder();
			for(int i : arr)
				ans.append(i + " ");
			System.out.println(ans);
		}
		file.close();
	}

}

class Segment implements Comparable<Segment> {
	int left, right, mid, length;
	public Segment(int l, int r) {
		left = l;
		right = r;
		mid = (left+right)/2;
		length = right-left;
	}
	public int compareTo(Segment s) {
		if(length == s.length) {
			return -(left - s.left);
		}
		return length-s.length;
	}
}