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:
//package constructingthearray;

import java.util.*;
import java.io.*;

public class constructingthearray {
	
	static PriorityQueue<int[]> q;
	
	public static void main(String[] args) throws IOException {
		BufferedReader fin = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(fin.readLine());
		StringBuilder fout = new StringBuilder();
		while(t-- > 0) {
			int n = Integer.parseInt(fin.readLine());
			int[] ans = new int[n];
			q = new PriorityQueue<int[]>((a, b) -> Comparator.compareIntArr(a, b));
			q.add(new int[] {n, 0});
			for(int i = 0; i < n; i++) {
//				for(int[] j : q) {
//					System.out.print(j[0] + " " + j[1] + ", ");
//				}
				//System.out.println();
				int[] next = q.poll();
				ans[(next[1] * 2 + next[0] - 1) / 2] = i + 1;
				//System.out.println("MID: " + ((next[1] * 2 + next[0] - 1) / 2));
				if(next[0] > 2) {
					q.add(new int[] {next[0] / 2 + (next[0] % 2 == 0? -1 : 0), next[1]});
					q.add(new int[] {next[0] / 2, (next[1] * 2 + next[0] - 1) / 2 + 1});
				}
				else if(next[0] == 2) {
					q.add(new int[] {1, next[1] + 1});
				}
				
			}
			for(int i = 0; i < n; i++) {
				fout.append(ans[i]).append(" ");
			}
			fout.append("\n");
		}
		System.out.print(fout);
	}
	
	static class Comparator {
		public static int compareIntArr(int[] a, int[] b) {
			if(Integer.compare(a[0], b[0]) != 0) {
				return -Integer.compare(a[0], b[0]);
			}
			return Integer.compare(a[1], b[1]);
		}
	}
}