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.*;
import java.util.regex.*;


public class Codechef
{
    static class SubArray
    {
        public int L;
        public int R;
        public int size;
        public SubArray(int l, int r)
        {
            L=l;
            R=r;
            size=(R-L)+1;
        }
    }
	public static void main (String[] args)
	{
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t--!=0)
		{
		    int n = sc.nextInt();
		    int arr[] = new int[n+1];
		    PriorityQueue<SubArray> pq = new PriorityQueue<SubArray>(5,new Comparator<SubArray>(){
		        public int compare(SubArray ob1, SubArray ob2)
		        {
		            if(ob1.size == ob2.size)
		                return ob1.L-ob2.L;
		            else
		                return ob2.size - ob1.size;
		        }
		    });
		    pq.add(new SubArray(1,n));
		    int k = 1;
		    for(int i = 1; i <= n; i++)
		    {
		        SubArray ok = (SubArray)pq.poll();
		        int left = ok.L;
		        int right = ok.R;
		        int mid = left + (right-left)/2;
		        arr[mid] = k;
		        k++;
		        pq.add(new SubArray(left,mid-1));
		        pq.add(new SubArray(mid+1,right));
		    }
		    for(int i = 1; i <= n; i++)
		    {
		        System.out.print(arr[i] + " ");
		    }
		    System.out.println();
		}
	}
}