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.*;
import java.util.*;

public class D
{
    public static void main(String[] args) throws Exception
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());
        
        for (int z = 0; z < t; z++)
        {
            int n = Integer.parseInt(br.readLine());
            int[] a = new int[n];
            PriorityQueue<Item> queue = new PriorityQueue<Item>();
            queue.add(new Item(0, n - 1));
            int k = 1;
            while (queue.size() > 0)
            {
                Item p = queue.remove();
                int m = (p.l + p.r) / 2;
                a[m] = k++;
                if (p.l == p.r)
                    continue;
                if (p.l <= m - 1)
                    queue.add(new Item(p.l, m - 1));
                if (m + 1 <= p.r)
                    queue.add(new Item(m + 1, p.r));
            }
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < n; i++)
            {
                sb.append(a[i]);
                sb.append(" ");
            }
            System.out.println(sb.toString());
        }
    }
}
class Item implements Comparable<Item>
{
    public int l;
    public int r;
    public int d;
    public Item(int l, int r)
    {
        this.l = l;
        this.r = r;
        this.d = r - l + 1;
    }
    public int compareTo(Item i)
    {
        if (this.d > i.d)
            return -1;
        if (this.d < i.d)
            return 1;
        if (this.l < i.l)
            return -1;
        if (this.l > i.l)
            return 1;
        return 0;
    }
}