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.Comparator;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Solution {
    private static final Comparator<int[]> CMP = new Comparator<int[]>() {
        @Override
        public int compare(int[] a, int[] b) {
            int cmp = Integer.compare(b[1] - b[0], a[1] - a[0]);
            if (cmp != 0) return cmp;

            return Integer.compare(a[0], b[0]);
        }
    };

    public static void main(String[] args) {
        final Scanner in = new Scanner(System.in);

        final int T = in.nextInt();

        for (int t = 0; t < T; t += 1) {
            final int n = in.nextInt();

            PriorityQueue<int[]> pq = new PriorityQueue<>(CMP);
            pq.offer(new int[] { 1, n });

            String[] a = new String[n];

            int i = 1;
            while (!pq.isEmpty()) {
                int[] lr = pq.poll();
                int l = lr[0];
                int r = lr[1];

                if (l > r) continue;

                int mid = ((r - l + 1) & 1) == 1 ? (l + r) / 2 : (l + r - 1) / 2;
                a[mid - 1] = String.valueOf(i++);

                pq.offer(new int[] {l, mid - 1});
                pq.offer(new int[] {mid + 1, r});
            }

            System.out.println(String.join(" ", a));
        }

    }
}