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.io.PrintStream;
import java.util.PriorityQueue;
import java.util.Scanner;

/**
 * Codeforces Problem 1353D
 */
public class Solution {

	public static void main(String args[]) {
		try (Scanner in = new Scanner(System.in);
				 PrintStream out = System.out;) {
			int t = in.nextInt();
			for (int i = 1; i <= t; i++) {
				String solution = solveNext(in);
//				out.println("Case #" + i + ": " + solution);
				out.println(solution);
				out.flush();
			}
		}
		System.exit(0);
	}

	public static String solveNext(Scanner in) {
		int n = in.nextInt();
		return new Solution(n).solve();
	}

	int n;

	public Solution(int n) {
		this.n = n;
	}

	private static class Segment implements Comparable<Segment> {

		int l;
		int r;
		int size;

		public Segment(int l, int r) {
			this.l = l;
			this.r = r;
			this.size = r - l + 1;
		}

		@Override
		public int compareTo(Segment o) {
			int comp = -Integer.compare(this.size, o.size);
			if(comp == 0){
				return Integer.compare(this.l, o.l);
			}
			return comp;
		}
	}

	public String solve() {
		int curr = 1;
		PriorityQueue<Segment> queue = new PriorityQueue<>();
		queue.add(new Segment(1, this.n));
		int[] arr = new int[this.n];
		while(!queue.isEmpty()){
			Segment next = queue.poll();
			int mid = (next.l + next.r) / 2;
			arr[mid - 1] = curr;
			curr++;
			if(mid > next.l){
				queue.add(new Segment(next.l, mid - 1));
			}
			if(mid < next.r){
				queue.add(new Segment(mid + 1, next.r));
			}
		}

		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < this.n; i++) {
			if (i > 0) {
				sb.append(" ");
			}
			sb.append(arr[i]);
		}
		return sb.toString();
	}
}