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 PYTHON3):

Solution:
def inverse_permutation(arr):
	inverse = [0] * len(arr)
	for i, (el, runlen) in enumerate(arr): inverse[el - 1] = (i, runlen)
	return inverse

solve_dict = {
	1 : [(1, 1)],
	2 : [(1, 2), (2, 1)],
}
def solve(n):
	if n in solve_dict: return solve_dict[n]
	answer = [None] * n
	k = (n - 1) // 2
	answer[k] = (1, n)
	lip = inverse_permutation(solve(k))
	rip = inverse_permutation(solve(n - k - 1))
	left_i = 0
	right_i = 0
	for i in range(2, n + 1):
		left_run_length = lip[left_i][1]
		right_run_length = rip[right_i][1]
		if left_run_length >= right_run_length:
			answer[lip[left_i][0]] = (i, lip[left_i][1])
			left_i += 1
			if left_i == k:
				while right_i < n - k - 1:
					i += 1
					answer[rip[right_i][0] + k + 1] = (i, rip[right_i][1])
					right_i += 1
				break
		else:
			answer[rip[right_i][0] + k + 1] = (i, rip[right_i][1])
			right_i += 1
			if right_i == n - k - 1:
				while left_i < k:
					i += 1
					answer[lip[left_i][0]] = (i, lip[left_i][1])
					left_i += 1
				break
	solve_dict[n] = answer
	return answer

t = int(input())
for _ in range(t):
	n = int(input())
	answer = solve(n)
	res = ''
	for x in answer: res += str(x[0]) + ' '
	res = res[:-1]
	print(res)