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 Abc {
	static class Node{
		int l, r;
		Node(int a, int b) {
			l = a;
			r = b;
		}
		int compareTo(Node a) {
			int size = size();
			int size_a = a.size();
			if(size > size_a) {
				return -1;
			} else if (size_a > size){
				return 1;
			} else {
				if(l < a.l) {
					return -1;
				} else {
					return 1;
				}
			}
		}
		int size() {
			return r - l + 1;
		}
		int getMid() {
			return (l + r) / 2;
		}
	}
	public static void main(String[] args) {
		Scanner in= new Scanner(System.in);
		int t = in.nextInt();
		while(t-->0) {
			int n = in.nextInt();
			PriorityQueue<Node> pq = new PriorityQueue<>(1, new Comparator<Node>(){
				public int compare(Node a, Node b) {
					return a.compareTo(b);
				}
			});
			pq.add(new Node(1, n));
			int ar[] = new int[n];
			int i = 1;
			while(!pq.isEmpty()) {
				Node node = pq.poll();
				//System.out.println("["+node.l+", "+node.r+"]");
				int index = node.getMid() - 1;
				ar[index] = i++;
				//System.out.println("placed "+ ar[index]+ " at "+(index+1));
				if(node.l != index + 1) {
					Node n1 = new Node(node.l, index);
					//System.out.println("Left: [" + node.l + ", " + index + "] Size: " + n1.size());
					pq.add(n1);
				}
				if(node.r != index + 1) {
					Node n2 = new Node(index + 2, node.r);
					//System.out.println("Right: [" + (index + 2) + ", " + node.r + "] Size: " + n2.size());
					pq.add(n2);
				}
			}
			for(int j = 0; j < n; j++) {
				System.out.print(ar[j]+ " ");
			}
			System.out.println();
		}
	}
}