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.Scanner;

public class Main {

    public static void main(String[] args) {
        
        Scanner sc = new Scanner(System.in);
        
        int T = Integer.parseInt(sc.next());
        StringBuilder sb = new StringBuilder();
        for(int t = 0; t < T; t++){
            int n = Integer.parseInt(sc.next());
            int[] a = new int[n];
            for(int i = 0; i < n; i++){
                a[i] = i+1;
            }
            
            PriorityQueue<Array> pq = new PriorityQueue<>();
            pq.add(new Array(a));
            int[] b = new int[n+1];
            int count = 1;
            while(!pq.isEmpty()){
                Array ar = pq.poll();
                if(ar.length == 1){
                    b[ar.a[0]] = count++;
                }else{
                    int mid = (ar.length -1) / 2;
                    b[ar.a[mid]] = count++;
                    if(ar.length > 2){
                        pq.add(new Array(Arrays.copyOfRange(ar.a, 0, mid)));
                    }
                    pq.add(new Array(Arrays.copyOfRange(ar.a, mid+1, ar.length)));
                }
            }
            
            for(int i = 1; i <= n; i++){
                sb.append(b[i]).append(" ");
            }
            sb.append(System.lineSeparator());
        }
        System.out.print(sb);
    }
    
    static class Array implements Comparable<Array>{
        int[] a;
        int length;
        int a0;
        public Array(int[] a) {
            this.a = a;
            this.length = a.length;
            a0 = a[0];
        }
        @Override
        public int compareTo(Array a) {
            int key1 = - Integer.compare(this.length, a.length);
            if(key1 != 0) return key1;
            return Integer.compare(this.a0, a.a0);
        }
    }
}