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.*;

    public class Main{
        static class Range{
            int l,r;
            int d;
            public Range(int l,int r) {this.l=l;this.r=r; d = r-l+1;}
            @Override
            public String toString(){return "["+this.l+", "+this.r+"]";}
        }
        public static void main(String... args) {
            Scanner sc = new Scanner(System.in);
            Comparator<Range> cmp = (Range r,Range q) ->{
                // r - l + 1  == r - l + 1
//              if(r.r-r.l + 1 == q.r - q.l +1) return -r.l + q.l;
//              return -(q.r - q.l + 1) + (r.r - r.l +1);
                if(r.d !=q.d) return q.d - r.d;
                return r.l - q.l;
            };
            PriorityQueue<Range> pq = new PriorityQueue<>(cmp);
            int tc = sc.nextInt();
            while(tc--!=0){
            int n = sc.nextInt();
            pq.add(new Range(0,n-1));

            int ar[] = new int[n];
            int k=1;
            while(!pq.isEmpty()){
                Range r = pq.remove();
                ar[(r.r+r.l)/2] = k++;
                if(r.l +1 == r.r){
                    pq.add(new Range(r.r,r.r));
                }
                else if(r.r!=r.l){
                    pq.add(new Range(r.l,((r.r+r.l)/2)-1));
                    pq.add(new Range(((r.r+r.l)/2)+1,r.r));
                }
            }
                System.out.print(ar[0]);
            for(int i = 1;i<ar.length;i++) System.out.print(" "+ar[i]);
                System.out.println();
            }
        }
    }