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 D {

	public static void main(String[] args) {
		new D();
	}

	public D() {
		Scanner sc = new Scanner(System.in);

		int amountT = sc.nextInt();
		for (int task = 0; task < amountT; task++) {
			PriorityQueue<Pair> pq = new PriorityQueue<Pair>();
			int n = sc.nextInt();
			pq.add(new Pair(0, n-1));
			int[] sol = new int[n];
			int index = 1;
			while(!pq.isEmpty()) {
				Pair p = pq.poll();
				int mid = -1;
				if ((p.end-p.start)%2 == 0) {
					mid = (p.start+p.end)/2;
				} else {
					mid = (p.start+p.end-1)/2;
				}
				if (mid-p.start > 0) {
					pq.add(new Pair(p.start, mid-1));
				}
				if (p.end-mid > 0) {
					pq.add(new Pair(mid+1, p.end));
				}
				sol[mid] = index;
				index++;
			}
			StringBuilder sb = new StringBuilder();
			for (int i = 0; i < n; i++) {
				if (i != 0) sb.append(' ');
				sb.append(sol[i]);
			}
			System.out.println(sb);
		}
		
		sc.close();
	}
	
	class Pair implements Comparable<Pair>{
		int start;
		int end;
		public Pair(int s, int e) {
			this.start = s;
			this.end = e;
		}
		public int compareTo(Pair p) {
			int dist = this.end-this.start;
			int pDist = p.end-p.start;
			if (dist == pDist) {
				return this.start-p.start;
			} else {
				return pDist-dist;
			}
		}
	}
}