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.*;

public class Main {
	static PriorityQueue<Integer> q;
	static HashMap<Integer, List<Integer> > hm;
	public static void mark(int[] arr, int l, int r){
		int val = r-l+1;
		
		arr[(l+r)/2]= val;
		int mid= (l+r)/2;
		if (hm.containsKey(val))hm.get(val).add(mid);
		else{
			List<Integer> list = new ArrayList<>();
			list.add(mid);
			hm.put(val,list);
		}

		if (l==r)return;
		q.add(r-mid);
		if (mid!=l) q.add(mid-l);	
		mark(arr, mid+1, r);

		if (mid!=l){
		
			mark(arr, l, mid - 1);

		}




	}

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int tests = sc.nextInt();
        for (int z=0;z<tests;z++) {
			hm = new HashMap<>();
			int n = sc.nextInt();
			q = new PriorityQueue<>(n, Collections.reverseOrder());
			int[] res = new int[n];
			int[] arr = new int[n];
			int[] cnt = new int[n+1];
			q.add(n);
			mark(arr, 0, n-1);
			int curr=1;

			while (q.size()!=0){
				int val = q.poll();
			
				if (cnt[val]!=0)continue;
				List<Integer> list = hm.get(val);
				Collections.sort(list);
				for (int i=0;i<list.size();i++){
					res[list.get(i)]=curr;
					curr++;
				}	
				cnt[val]++;
			}	
			for (int i=0;i<n-1;i++){
			
				System.out.print(res[i] + " ");
			}
			System.out.println(res[n-1]);




        }


    }
}