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.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.stream.Collectors;

public class Solution1353D {

    private static int cnt = 0;

    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int T = in.nextInt();
        PrintWriter writer = new PrintWriter(System.out);
        while (T-- > 0) {
            int n = in.nextInt();

            System.out.println(solve(n));
        }
    }

    private static String solve(int n) {
        int[] res = new int[n];
        cnt = 1;
        var pq = new PriorityQueue<int[]>((a, b) -> {
            if (a[1] - a[0] != b[1] - b[0]) {
                 return (b[1] - b[0]) - (a[1] - a[0]);
            } else {
                return a[0] - b[0];
            }
        });

        pq.add(new int[]{0, n - 1});

        while (!pq.isEmpty()) {
            var poll = pq.poll();
            int m = (poll[1] + poll[0])/2;
            res[m] = cnt++;
            if (poll[0] <= m - 1) pq.add(new int[]{poll[0], m - 1});
            if (m + 1 <= poll[1]) pq.add(new int[]{m + 1, poll[1]});
        }

        return Arrays.stream(res).mapToObj(String::valueOf).collect(Collectors.joining(" "));
    }

    static int[] toArray(String str) {
        return Arrays.stream(str.split(" ")).mapToInt(Integer::valueOf).toArray();
    }
}