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 ConstrArray {


	public static void main(String[] args) {

		Scanner scn = new Scanner(System.in);
		int t = scn.nextInt();

		while (t-- > 0) {

			int n = scn.nextInt();

			PriorityQueue<int[]> pq = new PriorityQueue<int[]>(new Comparator<int[]>() {

				@Override
				public int compare(int[] A, int[] B) {
					int da = dist(A);
					int db = dist(B);
					if (da == db) {
						return A[0] - B[0];
					}
					return db - da;
				}
			});

			pq.offer(new int[] {1, n});
			int[] res = new int[n];
			int i = 1;
			while (!pq.isEmpty()) {

				int[] rp = pq.poll();
				int mid = 0;
				if (dist(rp) % 2 == 0) {
					mid = (rp[1] + rp[0] - 1) / 2;
				} else {
					mid = (rp[1] + rp[0]) / 2;
				}

				res[mid - 1] = i++;
				if (mid - 1 >= rp[0])
					pq.offer(new int[] {rp[0], mid - 1});
				if (mid + 1 <= rp[1])
					pq.offer(new int[] {mid + 1, rp[1]});
			}

			for (int val : res)
				print(val + " ", false);

			print("", true);
		}

	}

	public static int dist(int[] A) {
		return A[1] - A[0] + 1;
	}

	public static <T> void print(T t, boolean ln) {
		System.out.print(t);
		if (ln) System.out.println();

	}

}