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.Comparator;
import java.util.PriorityQueue;
import java.util.Scanner;

public class ConstructingtheArray {

    public class Obj {

        int l;
        int r;
        int val;

        public Obj(int l, int r) {
            this.l = l;
            this.r = r;
        }
    }

    class ObjComparator implements Comparator<Obj> {

        public int compare(Obj a, Obj b) {
            if ((b.r - b.l) - (a.r - a.l) != 0) {
                return (b.r - b.l) - (a.r - a.l);
            } else {
                return a.l - b.l;
            }
        }
    }

    public int[] constructArray(int length) {
        int[] result = new int[length];
        //bfs
        int n = 1;
        PriorityQueue<Obj> pqueue = new PriorityQueue<>(new ObjComparator());
        pqueue.add(new Obj(0, length - 1));
        while (!pqueue.isEmpty()) {
            Obj current = pqueue.remove();
            int l = current.l;
            int r = current.r;
            if (l <= r) {
                int mid;
                if ((r - l + 1) % 2 == 0) {
                    mid = (l + r - 1) / 2;
                } else {
                    mid = (l + r) / 2;
                }
                result[mid] = n;
                n++;
                if (mid - 1 - l >= r - (mid + 1)) {
                    pqueue.add(new Obj(l, mid - 1));
                    pqueue.add(new Obj(mid + 1, r));
                } else {
                    pqueue.add(new Obj(mid + 1, r));
                    pqueue.add(new Obj(l, mid - 1));
                }
            }
        }
        return result;
    }

    public static void main(String args[]) {
        ConstructingtheArray cta = new ConstructingtheArray();
        Scanner sc = new Scanner(System.in);
        int testCase = sc.nextInt();
        for (int i = 1; i <= testCase; i++) {
            int length = sc.nextInt();
            int[] answer = cta.constructArray(length);
            for (int ans : answer) {
                System.out.print(ans + " ");
            }
            System.out.println();
        }
    }
}