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.PriorityQueue;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int t = s.nextInt();
		StringBuffer strbf = new StringBuffer();
		while(t-->0) {
			int n = s.nextInt();
			int count = 1;
			int[] arr = new int[n];
			PriorityQueue<Node> pq = new PriorityQueue<Node>((A, B) -> (A.len == B.len) ? (A.l - B.l) : (B.len - A.len));
			pq.add(new Node(0, n - 1));
			while(pq.size() > 0) {
				Node curr = pq.remove();
				if(curr.l > curr.r) {
					continue;
				}
				arr[(curr.l + curr.r) / 2] = count++;
				int mid = (curr.l + curr.r) / 2; 
				pq.add(new Node(curr.l, mid - 1));
				pq.add(new Node(mid + 1, curr.r));
			}
			for(int i = 0; i < n; i++) {
				strbf.append(arr[i] + " ");
			}
			strbf.append(System.lineSeparator());
		}
		System.out.println(strbf.toString());
	}

}

class Node {
	int l, r, len;
	Node(int l, int r) {
		this.l = l;
		this.r = r;
		len = r - l + 1;
	}
}