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 CF1353D extends PrintWriter {
	CF1353D() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1353D o = new CF1353D(); o.main(); o.flush();
	}

	static class V {
		int i, n;
		V(int i, int n) {
			this.i = i; this.n = n;
		}
	}
	V[] qu; int cnt;
	void solve(int i, int n) {
		if (n == 0)
			return;
		qu[cnt++] = new V(i, n);
		int nl = (n - 1) / 2;
		solve(i, nl);
		solve(i + nl + 1, n - nl - 1);
	}
	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			qu = new V[n]; cnt = 0;
			solve(0, n);
			Arrays.sort(qu, (u, v) -> u.n != v.n ? v.n - u.n : u.i - v.i);
			int[] aa = new int[n];
			for (int i = 0; i < n; i++) {
				V v = qu[i];
				aa[v.i + (v.n - 1) / 2] = i + 1;
			}
			for (int i = 0; i < n; i++)
				print(aa[i] + " ");
			println();
		}
	}
}