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.Comparator;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t != 0) {
			int n = sc.nextInt();
			solve(n);
			t--;
		}
	}

	public static void solve(int n) {
		int[] a = new int[n + 1];

		PriorityQueue<Pair> queue = new PriorityQueue<Pair>(n, new PairComparator());
		queue.add(new Pair(1, n));
		
		int action = 1;
		
		while(!queue.isEmpty()) {
			Pair currPair = queue.poll();
			int length = currPair.b - currPair.a + 1;
			
			int left = currPair.a, right = currPair.b;
			int index = 0;
			
			
			if(length % 2 != 0) {
				index = (left + right) / 2;
			} else {
				index = (left + right - 1) / 2;
			}
			a[index] = action++;
			
			if(currPair.a <= (index - 1)) {
				queue.add(new Pair(currPair.a, index - 1));
			}
			
			if(currPair.b >= (index + 1)) {
				queue.add(new Pair(index + 1, currPair.b));
			}
			
		}
		
		for(int i = 1; i <= n;  i++) {
			System.out.print(a[i] + " ");
		}
		System.out.println();
	}

}

class Pair {
	public int a;
	public int b;
	public Pair(int a, int b) {
		this.a = a;
		this.b = b;
	}
}

class PairComparator implements Comparator<Pair> {

	public int compare(Pair o1, Pair o2) {
		int l1 = o1.b - o1.a;
		int l2 = o2.b - o2.a;
		if(l2 > l1) {
			return 1;
		} else if(l1 > l2) {
			return -1;
		} else {
			if(o1.a < o2.a) {
				return -1;
			} else {
				return 1;
			}
		}
	}

}