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 D1353 {
    static int arr[];

    public static void main(String args[]) throws IOException {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int z = 1; z <= t; z++) {
            int n = sc.nextInt();
            PriorityQueue<pair> pq = new PriorityQueue<>(new Comparator<pair>() {
                public int compare(pair ob1, pair ob2) {
                    int r = Integer.compare(ob2.diff, ob1.diff);
                    if (r == 0)
                        return Integer.compare(ob1.left, ob2.left);

                    return r;
                }
            });
            int arr[] = new int[n];
            pq.add(new pair(0, n - 1));
            for (int i = 1; i <= n; i++) {
                pair ob = pq.poll();
                int mid = (ob.left + ob.right) / 2;
                arr[mid] = i;
                if (ob.left != ob.right) {
                    pq.add(new pair(ob.left, mid - 1));
                    pq.add(new pair(mid + 1, ob.right));
                }
            }
            for (int i = 0; i < n; i++) {
                System.out.print(arr[i] + " ");
            }
            System.out.println();
        }
    }

    static class pair {
        int left, right, diff;

        public pair(int l, int r) {
            left = l;
            right = r;
            diff = right - left + 1;
        }
    }
}