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 cf1353_Div3_D {
	
	public static class Seg implements Comparable <Seg>{
		int l;
		int r;
		int len;
		int len2;
		
		public Seg(int x, int y) {
			l = x;
			r = y;
			len = r - l;
			len2 = r + l;
		}
		
		public int compareTo(Seg other) {
			if (other.len == len)
				return l - other.l;
			return other.len - len;
		}
	}
	
	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		for ( ; t > 0; t--) {
			int n = in.nextInt();
			int[] arr = new int[n];
			
			PriorityQueue<Seg> q = new PriorityQueue<Seg>();
			q.add(new Seg(0, n - 1));
			
			int i = 1;
			while (!q.isEmpty() && i <= n) {
				Seg s = q.poll();
			
				int pos = 0;
				//System.out.println(s.len);
				if ((s.len + 1) % 2 == 0) {
					pos = (s.len2 - 1) / 2;
				}
				else {
					pos = s.len2 / 2;
				}
				arr[pos] = i;
				// System.out.println(s.l + " " + s.r + " " + pos);
				i++;
				if (s.l == s.r) continue;
				if (pos - 1 >= 0)
					q.add(new Seg(s.l, pos - 1));
				if (pos + 1 < n)
					q.add(new Seg(pos + 1, s.r));
			}
			
			for (i = 0; i < arr.length; i++) {
				System.out.print(arr[i] + " ");
			}
			System.out.println();
			
		}
	}
}