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.PriorityQueue;
import java.util.Scanner;

public class D {

	static class Segment implements Comparable<Segment> {
		
		public int s, e;
		
		public Segment(int si, int ei) {
			s = si;
			e = ei;
		}

		@Override
		public int compareTo(Segment b) {
			if (e - s - (b.e - b.s) == 0) {
				if (s < b.s) {
					return -1;
				}
				return 1;
			} else {
				return (b.e - b.s) - (e - s);
			}
		}
		
	}
	
	public static void main(String[] args) {
		
		Scanner s = new Scanner(System.in);
		int t = s.nextInt();
		for (int z = 0; z < t; z++) {
			int n = s.nextInt();
			PriorityQueue<Segment> lengths = new PriorityQueue<Segment>();
			lengths.add(new Segment(0, n - 1));
			int[] arr = new int[n];
			int op = 1;
			while (!lengths.isEmpty()) {
				Segment c = lengths.poll();
				int length = c.e - c.s + 1;
				if (length % 2 == 0) {
					int index = (c.s + c.e - 1) / 2;
					arr[index] = op++;
					if (length == 2) {
						lengths.add(new Segment(index + 1, c.e));
					} else {
						lengths.add(new Segment(c.s, index - 1));
						lengths.add(new Segment(index + 1, c.e));
					}
				} else {
					int index = (c.s + c.e) / 2;
					arr[index] = op++;
					if (length != 1) {
						lengths.add(new Segment(c.s, index - 1));
						lengths.add(new Segment(index + 1, c.e));
					}
				}
			}
			StringBuilder ans = new StringBuilder("");
			for (int i = 0; i < n; i++) {
				ans.append(arr[i]);
				if (i != n - 1) {
					ans.append(" ");
				}
			}
			System.out.println(ans);
		}
		
	}
	
}