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.*;
import java.util.*;
public class D
{
    static BufferedReader br;
    static BufferedWriter bw;
    static int A[];
    static void solve() throws Exception
    {
        int n = Integer.parseInt(br.readLine());
        A = new int[n+1];
        PriorityQueue<T> q = new PriorityQueue<T>();
        q.add(new T(n,1,n));
        int value = 1;
        while(!q.isEmpty())
        {
            T temp = q.remove();
            int l = temp.l;
            int r = temp.r;
            int mid;
            if((r-l+1)%2==1)
            {
                mid = (l+r)/2;
                if(l<=mid-1)
                q.add(new T(mid-l,l,mid-1));
                if(mid+1<=r)
                q.add(new T(r-mid,mid+1,r));
            }
            else
            {
                mid = (l+r-1)/2;
                if(l<=mid-1)
                q.add(new T(mid-l,l,mid-1));
                if(mid+1<=r)
                q.add(new T(r-mid,mid+1,r));
            }
            A[mid] = value;
            value++;
        }
        for(int i=1;i<=n;i++)
        bw.write(Integer.toString(A[i])+" ");
        bw.newLine();
    }
    public static void main(String[] args)throws Exception {
        br = new BufferedReader(new InputStreamReader(System.in));
        bw = new BufferedWriter(new OutputStreamWriter(System.out));
        int t=Integer.parseInt(br.readLine());
        for(int i=0;i<t;i++)
        {
            solve();
        }
        br.close();
        bw.close();
    }
}
class T implements Comparable<T>
{
    int size,l,r;
    T(int size,int l,int r)
    {
        this.size = size;
        this.l = l;
        this.r = r;
    }
    public int compareTo(T other)
    {
        if(this.size>other.size)
        return -1;
        else if(this.size<other.size)
        return 1;
        else
        {
            if(this.l<other.l)
            return -1;
            else if(this.l>other.l)
            return 1;
            return 0;
        }
    }
}