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.BufferedWriter;
import java.io.IOException;
import java.io.OutputStreamWriter;
import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.HashSet;
import java.util.HashMap;
import java.util.PriorityQueue;
import java.util.Scanner;



public class CF642n3 {
    public static void main(String[] args) throws IOException {
        Scanner scan = new Scanner(System.in);
        BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
        //log.write() //log.flush()
        int t = scan.nextInt();
        int count = 1;
        while (t-- > 0) {
            int n = scan.nextInt();

            int[] arr = new int[n];

//            Comparator<Pair> comparator = new PairComparator();
            PriorityQueue<Pair> queue = new PriorityQueue<>(n, (x,y)->{
                if(x.len < y.len){
                    return 1;
                }
                else if(y.len<x.len){
                    return -1;
                }
                else{
                    if(x.ind<y.ind){
                        return -1;
                    }
                    else if(y.ind<x.ind){
                        return 1;
                    }
                    else{
                        return 0;
                    }
                }
            });

            Pair first = null;
            if(n%2==0) {
                first = new Pair(n, 0);
            }
            else{
                first = new Pair(n, 0);
            }
            queue.add(first);

            int counter = 1;
            while(queue.size() !=0){
                Pair rem = queue.remove();
                int L = rem.ind;
                int R = rem.ind + rem.len -1;
                if (L > R) {
                    continue;
                }
                int mid = (L + R) / 2;
                arr[mid] = counter;

                if (L != R) {
                    //split into 2
                    if (R - L == 1) { // 00
                        queue.add(new Pair(1, mid+1));
                    } else { //00000
                        queue.add(new Pair(mid-L, L));
                        queue.add(new Pair(R-mid, mid+1));
                    }
                }
//                if(rem.len ==1){
//                    arr[rem.ind] = counter;
//                }
//                else if(rem.len%2==0){
//                    arr[rem.ind + (rem.len/2) -1] = counter;
//                    int mid = rem.ind + (rem.len/2) -1;
//                    Pair f = new Pair(mid, rem.ind);
//                    Pair sec = new Pair(mid+1, mid+1);
//                    queue.add(f);
//                    queue.add(sec);
//                }
//                else {
//                    arr[rem.ind + (rem.len/2)] = counter;
//                    int mid = rem.ind + (rem.len/2);
//                    Pair f = new Pair(mid, rem.ind);
//                    Pair sec = new Pair(mid, mid+1);
//                    queue.add(f);
//                    queue.add(sec);
//                }
                counter++;
            }


            for(int num: arr) {
                log.write("" + num + " ");
            }
            log.write("\n");

            count++;
        }
        log.flush();


    }
    static class Pair{
        int len;
        int ind;
        Pair (int len, int ind){this.len = len; this.ind=ind; }
    }
}