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.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.HashMap;
import java.util.PriorityQueue;

public class ConstructTheArray {
    public static void main(String[] args) throws IOException{
        BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(r.readLine());
        while(t -- > 0){
            int n = Integer.parseInt(r.readLine());
            PriorityQueue <Indexes>queue = new PriorityQueue<>(new compp());
            queue.add(new Indexes(0, n));
            int[]sol = new int[n];
            int current = 1;
            for(int i = n; i > 0; i --){
                Indexes now = queue.remove();
                int left = now.left, right = now.right;
                int mid = left + (right - left) / 2 - (1 - (right - left) % 2);
                sol[mid] = current ++;
                if(left < mid)
                    queue.add(new Indexes(left, mid));
                if(right > mid + 1)
                    queue.add(new Indexes(mid + 1, right));
                 
            }
            StringBuilder b = new StringBuilder();
            for(int i :sol){
                b.append(i);
                b.append(" ");
            }
            System.out.println(b);
        }
    }
        
}
class compp implements Comparator<Indexes>{

    @Override
    public int compare(Indexes n1, Indexes n2) {
        if(n1.right - n1.left !=n2.right - n2.left)
            return Integer.compare(n2.right - n2.left,n1.right - n1.left);
        return Integer.compare(n1.left, n2.left);
    }

}
class Indexes{
    int left, right;

    Indexes(int left, int right){
        this.left = left;
        this.right = right;
    }

}