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 {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		for(int i = 0; i < t; i++) {
			int n = sc.nextInt();
			int l = 1, r = n;
			Queue<int[]> q = new PriorityQueue<>((a,b) -> b[2] == a[2] ? a[0] - b[0] : b[2] - a[2]);
			int[] res = new int[n + 1];
			q.offer(new int[] {l, r, r - l});
			for(int j = 1; j <= n; j++) {
				int[] curr = q.poll();
				int left = curr[0];
				int right = curr[1];
				if(left == right) {
					res[left] = j;
					continue;
				}
				int mid = 0;
				if((right - left + 1) % 2 == 1) {
					mid = (left + right) / 2;
					res[mid] = j;
				}else {
					mid = (left + right - 1) / 2;
					res[mid] = j;
				}
				q.offer(new int[] {left, mid - 1, mid - left - 1});
				q.offer(new int[] {mid + 1, right, right - mid - 1});
			}
			for(int j = 1; j <= n; j++) {
				System.out.print(res[j] + " ");
			}
			System.out.println();
		}
	}
}