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.Arrays;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;

public class ConstructingtheArray {

  private static class Pair {
    int l;
    int r;

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

  private static int findMid(int l, int r) {
    int size = r - l + 1;
    if (size % 2 != 0) {
      return (l + r) / 2;
    } else {
      return (l + r - 1) / 2;
    }
  }

  //  private static int doProcess(int[] a, int l, int r, int step) {
  //    if (l > r) return step;
  //    int size = r - l + 1;
  //    int mid;
  //    if (size % 2 != 0) {
  //      mid = (l + r) / 2;
  //    } else {
  //      mid = (l + r - 1) / 2;
  //    }
  //    a[mid] = step;
  //    step = doProcess(a, l, mid - 1, step + 1);
  //    step = doProcess(a, mid + 1, r, step);
  //    return step;
  //  }

  private static void doProcess(Queue<Pair> q, int l, int r) {
    if (l > r) return;
    q.add(new Pair(l, r));
    int mid = findMid(l, r);
    doProcess(q, l, mid - 1);
    doProcess(q, mid + 1, r);
  }

  private static void process(int n) {
    int[] a = new int[n];
    Arrays.fill(a, 0);
    Queue<Pair> q =
        new PriorityQueue<>(
            n,
            (p1, p2) -> {
              int s1 = p1.r - p1.l;
              int s2 = p2.r - p2.l;
              if (s1 == s2) return p1.l - p2.l;
              else return s2 - s1;
            });
    doProcess(q, 0, n - 1);
    int step = 1;
    while (!q.isEmpty()) {
      Pair p = q.poll();
      int idx = findMid(p.l, p.r);
      a[idx] = step;
      step++;
    }
    for (int i : a) {
      System.out.print(i + " ");
    }
  }

  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int t = in.nextInt();
    for (int i = 0; i < t; i++) {
      int n = in.nextInt();
      process(n);
      System.out.println();
    }
  }
}