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.*;
import java.io.*;

public class Solution {
  public void solve(FastReader in, PrintWriter pw) {
    int t = in.nextInt();

    while (t-- > 0) {
      int n = in.nextInt();
      int[] arr = new int[n];
      int c = 1;
      PriorityQueue<Pair> pq = new PriorityQueue<>();
      pq.offer(new Pair(0, n - 1));

      while (!pq.isEmpty()) {
        Pair curr = pq.poll();
        int mid = curr.left + (curr.right - curr.left) / 2;
        arr[mid] = c++;
        if (mid - 1 >= curr.left) pq.offer(new Pair(curr.left, mid - 1));
        if (mid + 1 <= curr.right) pq.offer(new Pair(mid + 1, curr.right));
      }

      for (int x : arr) pw.print(x + " ");
      pw.print("\n");
    }
  }

  public static void main(String[] args) throws IOException {
    FastReader in = new FastReader();
    PrintWriter pw = new PrintWriter(System.out);
    new Solution().solve(in, pw);
    pw.flush();
    pw.close();
  }

  private void ln(String str) {
    System.out.println(str);
  }
}

class Pair implements Comparable<Pair> {
  public int left, right, len;
  public Pair(int left, int right) {
    this.left = left;
    this.right = right;
    len = right - left;
  }

  @Override
  public int compareTo(Pair o) {
    if (this.len == o.len) return this.left - o.left;
    return o.len - this.len;
  }
}

class FastReader
{
  BufferedReader br;
  StringTokenizer st;

  public FastReader()
  {
    br=new BufferedReader(new InputStreamReader(System.in));
  }

  public String next()
  {
    if(st==null || !st.hasMoreElements())
    {
      try {
        st=new StringTokenizer(br.readLine());
      } catch(Exception e) {
        throw new RuntimeException(e);
      }
    }
    return st.nextToken();
  }

  public int nextInt()
  {
    return Integer.parseInt(next());
  }

  public long nextLong()
  {
    return Long.parseLong(next());
  }
}