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.Arrays;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Codeforces {
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		while(t-->0){
			int n= in.nextInt();
			
            int[] arr = new int[n+1];

            PriorityQueue<int[]> pq = new PriorityQueue<>((a,b) ->{
            	if(b[0]==a[0]){
            		return a[1]-b[1];
            	} else return b[0]-a[0]; 
            });
            
            pq.add(new int[]{n, 1, n});
            int cont =1;
            while(!pq.isEmpty()) {
            	int[] tt = pq.poll();
            	//System.out.println(":" +Arrays.toString(tt));
                int sum = tt[1] + tt[2];
                int index = sum%2==0 ? sum/2 : (sum-1)/2;
                arr[index] = cont;
                int leftPortion = index - 1 - tt[1] + 1;
                int rightPortion = tt[2]-(index+1) + 1;
                if(leftPortion>0)
                    pq.add(new int[]{leftPortion, tt[1], index - 1});
                if(rightPortion>0)
                    pq.add(new int[]{rightPortion, index + 1, tt[2]});
                cont++;
            }
		
            for(int i =1; i<=n; i++)
            	System.out.print(arr[i]+ " ");
            
            System.out.println();
		}
		
     }
}