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 D {
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0){
			int n=sc.nextInt();
			int arr[]=new int [n+1];
			int count=1;
			LinkedList<Long> list[]=new LinkedList[n+1];
			for(int i=0;i<=n;i++){
				list[i]=new LinkedList<Long>();
			}
			list[n].add((long) (1*1000000+n));
//			LinkedList<Long> big=new LinkedList<Long>();
//			big.add((long) (1*1000000+n));
//			LinkedList<Long> smail=new LinkedList<Long>();
			
			int blen=1,slen=0;
			long tt;
			long l,r,mid;
			
			
			for(int i=n;i>=0;i--){
				Collections.sort(list[i]);
				int len=list[i].size();
				for(int j=0;j<len;j++){
					tt=list[i].get(0);
					l=(int) (tt/1000000);
					r=(int) (tt%1000000);
					if(l>r)continue;
					mid=(l+r)/2;
					arr[(int) mid]=count;
					count++;
					
					if(l>=r){
						
					}else if(r-(mid+1)>mid-1-l){
						list[(int) (r-mid)].add((mid+1)*1000000+r);
						list[(int) (mid-l)].add(l*1000000+mid-1);
					}else{
						list[(int) (mid-l)].add(l*1000000+mid-1);
						list[(int) (r-mid)].add((mid+1)*1000000+r);
					}
//					System.out.println(i+" "+j+" "+list[i].size());
					list[i].remove(0);
				}
			}
			
//			
//			while(count<=n){
////				System.out.println(Arrays.toString(arr));
//				Collections.sort(big);
//				Collections.sort(smail);
//				blen=big.size();
//				slen=smail.size();
////				System.out.println(blen+" "+slen+" ");
//				for(int i=0;i<blen;i++){
//					tt=big.get(0);
//					l=(int) (tt/1000000);
//					r=(int) (tt%1000000);
//					if(l>r)continue;
//					mid=(l+r)/2;
//					arr[(int) mid]=count;
//					count++;
//					
//					if(l>=r){
//						
//					}else if(r-(mid+1)>mid-1-l){
//						big.add((mid+1)*1000000+r);
//						smail.add(l*1000000+mid-1);
//					}else{
//						smail.add(l*1000000+mid-1);
//						smail.add((mid+1)*1000000+r);
//					}
//					big.remove(0);
//				}
//				for(int i=0;i<slen;i++){
//					tt=smail.get(0);
////					System.out.println(tt);
//					l=(int) (tt/1000000);
//					r=(int) (tt%1000000);
//					if(l>r)continue;
//					mid=(l+r)/2;
//					arr[(int) mid]=count;
//					count++;
//					if(l>=r){
//						
//					}else if(r-(mid+1)>mid-1-l){
//						if(mid+1<=r)
//						big.add((mid+1)*1000000+r);
//						if(l<=mid-1)
//						smail.add(l*1000000+mid-1);
//					}else{
//						if(l<=mid-1)
//						smail.add(l*1000000+mid-1);
//						if(mid+1<=r)
//						smail.add((mid+1)*1000000+r);
//					}
//					smail.remove(0);
//				}
//			}
			StringBuilder ans=new StringBuilder();
			for(int i=1;i<=n;i++){
				ans.append((i==1?"":" ")+arr[i]);
			}
			System.out.println(ans);

		}
	}
}