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.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.Scanner;


public class Task642D {


    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int t = in.nextInt(); // Scanner has functions to read ints, longs, strings, chars, etc.
        int n;
        int a[] = new int[200010];
        PriorityQueue<Node> pq = new PriorityQueue<>((o1, o2) -> {
            if (o1.diff > o2.diff || o1.diff == o2.diff && o1.start < o2.start)
                return -1;
            return 1;
        });

        for (int i = 1; i <= t; ++i) {
            n = in.nextInt();
            int s = 1;
            int middle = (n + 1) / 2;
            a[middle] = s++;
            if (middle -2 >= 0)
                pq.add(new Node(1, middle - 1, middle - 2));
            if (n - middle-1 >= 0)
                pq.add(new Node(middle + 1, n, n - middle-1 ));
            while (pq.size() > 0) {
                final Node poll = pq.poll();
                middle = (poll.start + poll.end) / 2;
                a[middle] = s++;
                if (middle -1- poll.start >= 0) {
                    pq.add(new Node(poll.start, middle - 1, middle -1- poll.start));
                }
                if (-middle + poll.end > 0) {
                    pq.add(new Node(middle + 1, poll.end,  -1-middle + poll.end));
                }
            }
            for (int j = 1; j <= n; j++)
                System.out.print(a[j] + " ");
            System.out.println();
            Arrays.fill(a,0,n,0);


        }

    }

    static class Node {
        int start;
        int end;
        int diff;

        public Node(int start, int end, int diff) {
            this.start = start;
            this.end = end;
            this.diff = diff;
        }
    }
}