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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class App {

	static class Node {
		int size;
		int start;
		int end;

		public Node(int size, int start, int end) {
			super();
			this.size = size;
			this.start = start;
			this.end = end;
		}

		@Override
		public String toString() {
			return "(" + start + "," + end + ":" + size + ")";
		}
	}

	public static void main(String[] args) {
		Comparator<Node> comparator = new Comparator<Node>() {
			@Override
			public int compare(Node l, Node r) {
				if (l.size == r.size) return l.start - r.start;
				return r.size - l.size;
			}
		};

		FastReader fr = new FastReader();
		int testCases = fr.nextInt();
		for (int tc = 0; tc < testCases; tc++) {
			int size = fr.nextInt();
			PriorityQueue<Node> pq = new PriorityQueue<>(5, comparator);
			pq.add(new Node(size, 1, size));
			int[] arr = new int[size];
			int action = 1;
			while (pq.size() > 0) {
				Node node = pq.poll();
				int index = ((node.end - node.start + 1) & 1) == 1 ? (node.start + node.end) / 2
						: (node.start + node.end - 1) / 2;
				arr[index - 1] = action++;

				Node n1 = new Node(index - node.start, node.start, index - 1);
				Node n2 = new Node(node.end - index, index + 1, node.end);
				if (n1.size > 0) pq.add(n1);
				if (n2.size > 0) pq.add(n2);
			}

			StringBuilder sb = new StringBuilder();
			sb.append(arr[0]);
			for (int i = 1; i < size; i++) sb.append(" ").append(arr[i]);
			String output = sb.toString();
			System.out.println(output);
		}
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				}
				catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		String nextLine() {
			String s = "";
			try {
				s = br.readLine();
			}
			catch (IOException e) {
				e.printStackTrace();
			}
			return s;
		}
	}
}