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 D {
	
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(br.readLine());
		StringBuilder answer = new StringBuilder();
		for(int tc = 1; tc <= t; ++tc) {
			int n = Integer.parseInt(br.readLine());
			int[] arr = new int[n+1];
			Queue<int[]> q = new PriorityQueue<int[]>(new Comparator<int[]>() {
				@Override
				public int compare(int[] o1, int[] o2) {
					if(o1[1]-o1[0] > o2[1]-o2[0]) {
						return -1;
					}else if(o1[1]-o1[0] < o2[1]-o2[0]) {
						return 1;
					}
					return Integer.compare(o1[0], o2[0]);
				}
			});
			q.add(new int[] {1, n});
			for(int i = 1; i <= n; ++i) {
				int mid;
				int[] lra = q.remove();
				int l = lra[0];
				int r = lra[1];
				if((r-l+1)%2==1) {
					mid = (l+r)/2;
				}else {
					mid = (l+r-1)/2;
				}
				arr[mid] = i;
				int ll = l;
				int lr = mid-1;
				int rl = mid+1;
				int rr = r;
				if(rr-rl>lr-ll) {
					q.add(new int[] {rl, rr});
					q.add(new int[] {ll, lr});
				}else {
					q.add(new int[] {ll, lr});
					q.add(new int[] {rl, rr});
				}
			}
			StringBuilder sb = new StringBuilder();
			for(int i = 1; i <= n; ++i) {
				sb.append(arr[i]).append(' ');
			}
			sb.append('\n');
			answer.append(sb);
		}
		System.out.println(answer);
	}
}