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.*;
import java.math.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
	public static void main (String[] args) {
		Scanner sc=new Scanner(System.in);
	        int t = sc.nextInt();
	        while (t-- > 0) {
	            int n = sc.nextInt();
	            int k = 1;
	            int[] arr = new int[n];
	            PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> {//重写，划分的条件，得到题中的两个数的数组
	                int len1 = a[1]-a[0]+1, len2 = b[1]-b[0]+1;
	                if (len1 != len2) return (len2-len1); 
	                return (a[0]-b[0]);});
	 
	            pq.add(new int[]{0, n-1});//直接划分得到想要的结果
	            while (!pq.isEmpty()) {//pq不为空时
	                int[] cur = pq.poll();//检索并删除此队列的头	                
	                int len = cur[1] - cur[0] + 1;
	                int mid = (cur[1]+cur[0])/2;//获得下标
	                arr[mid] = k++;
	                if (len == 1) continue;//len为1时没有了
	                if ((mid-1) >= cur[0])
	                    pq.offer(new int[]{cur[0], mid-1});
	                if ((mid+1) <= cur[1])
	                    pq.offer(new int[]{mid+1, cur[1]});
	            }
	            for (int elem: arr) System.out.print(elem + " " );
	            System.out.println();
	        }
	    }
}