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 ConstructingTheArray {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int T = sc.nextInt();
    for (int i = 0; i < T; ++i) {
      int x = sc.nextInt();
      System.out.println(solve(x));
    }
  }

  static class Diff {
    int l;
    int r;

    public Diff(int l, int r) {
      this.l = l;
      this.r = r;
    }
  }

  private static String solve(int x) {
    int[] arr = new int[x + 1];
    StringBuilder s = new StringBuilder();
    PriorityQueue<Diff> p =
        new PriorityQueue<>(
            (o1, o2) ->
                (o2.r - o2.l + 1) - (o1.r - o1.l + 1) != 0
                    ? (o2.r - o2.l + 1) - (o1.r - o1.l + 1)
                    : o1.l - o2.l);
    int count = 1;
    Diff d = new Diff(1, x);
    p.add(d);
    while (!p.isEmpty() && count <= x) {
      Diff d1 = p.poll();
      if ((d1.r - d1.l + 1) % 2 != 0) {
        arr[(d1.r + d1.l) / 2] = count++;
        Diff d2 = new Diff(((d1.r + d1.l) / 2) + 1, d1.r);
        Diff d3 = new Diff(d1.l, ((d1.r + d1.l) / 2) - 1);
        if (d2.r >= d2.l) {
          p.add(d2);
        }
        if (d3.r >= d3.l) {
          p.add(d3);
        }
      } else {
        arr[(d1.r + d1.l - 1) / 2] = count++;
        Diff d2 = new Diff(((d1.r + d1.l - 1) / 2) + 1, d1.r);
        Diff d3 = new Diff(d1.l, ((d1.r + d1.l - 1) / 2) - 1);
        if (d2.r >= d2.l) {
          p.add(d2);
        }
        if (d3.r >= d3.l) {
          p.add(d3);
        }
      }
    }
    for (int i = 1; i <= x; i++) {
      s.append(arr[i]).append(" ");
    }
    return s.toString();
  }
}