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:

//1353D
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.PriorityQueue;

public class Constructing_the_Array {
    public static void main(String[] args) throws IOException {

        BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
        String test_numS=bf.readLine();
        int test_num=Integer.parseInt(test_numS);
        while(test_num>0){//当测试数量test_num大于零时循环输入测试数据
            String nS=bf.readLine();
            int n=Integer.parseInt(nS);
            int a[]=solve(n);
            for (int i = 0; i <n ; i++) {
                System.out.print(a[i]+" ");
            }
            System.out.println();
            test_num--;
        }
    }
    public static  int[] solve(int n){
        int ans[]=new int[n];
        PriorityQueue<PQItem> pq=new PriorityQueue<>();
        pq.add(new PQItem(n,1,n));
        for (int i = 1; i <=n ; i++) {
            PQItem p=pq.poll();
            int location=((p.right_No-p.left_No+1)%2==1)?(p.left_No+p.right_No)/2-1:(p.left_No+p.right_No-1)/2-1;
            ans[location]=i;
            pq.add(new PQItem(location-p.left_No+1,p.left_No,location));
            pq.add(new PQItem(p.right_No-location-1,location+2,p.right_No));
        }
        return ans;
    }
}
class PQItem implements Comparable<PQItem> {
    public  int zero_Num_len,left_No,right_No;
    public PQItem(int zero,int left,int right){
        zero_Num_len=zero;
        left_No=left;
        right_No=right;
    }
    public int compareTo(PQItem o) {
        if (this.zero_Num_len!=o.zero_Num_len)
            return o.zero_Num_len-this.zero_Num_len;
        else if (this.left_No!=o.left_No)
            return this.left_No-o.left_No;
        else return this.right_No-o.right_No;
    }
}