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.util.*;
 
public class Codechef{
    public static void main(String[] args){
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        try{
            int testCases = Integer.parseInt(br.readLine());
            
            for(int i=0; i<testCases; i++){
            	int n = Integer.parseInt(br.readLine());
            	
            	PriorityQueue<Node> q = new PriorityQueue<>();
            	q.add(new Node(0, n-1));
            	int[] ans = new int[n];
            	
            	int operation = 1;
            	
            	while(!q.isEmpty()) {
            		Node curr = q.remove();
            		if((curr.right-curr.left+1 % 2) == 0) {
            			int index = (curr.left+curr.right-1)/2;
            			ans[index] = operation;
            			operation++;
            			if(index != curr.left) q.add(new Node(curr.left, index-1));
            			if(index != curr.right) q.add(new Node(index+1, curr.right));
            		}
            		else {
            			int index = (curr.left+curr.right)/2;
            			ans[index] = operation;
            			operation++;
            			if(index != curr.left) q.add(new Node(curr.left, index-1));
            			if(index != curr.right) q.add(new Node(index+1, curr.right));	
            		}
            	}
            	
            	StringBuilder sb = new StringBuilder("");
            	for(int j=0;j<n;j++) {
            		sb.append(ans[j] + " ");
            	}
            	sb.substring(0, sb.length()-1);
            	System.out.println(sb);
            }
        }
        catch(Exception e){
            e.printStackTrace();
        }
    }
    
}

class Node implements Comparable<Node>{
	int left;
	int right;
	public Node(int left, int right) {
		this.left = left;
		this.right = right;
	}
	
	public int compareTo(Node other) {
		if(other.right - other.left < this.right - this.left) return -1;
		if(other.right - other.left > this.right - this.left) return 1;
		else return Integer.compare(this.left, other.left);
	}
}