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 ri(): return int(input())
def ria(): return list(map(int, input().split()))
def ia_to_s(a): return ' '.join([str(s) for s in a])


import heapq


class Range(object):
    def __init__(self, left, right):
        self.left = left
        self.right = right

    def __repr__(self):
        return f'Range: [{self.left},{self.right})'

    def __lt__(self, other):
        this_len = self.right-self.left
        other_len = other.right-other.left

        if this_len == other_len:
            return self.left < other.left
        else:
            return this_len > other_len


def solve(n):
    a = [0] * n
    ranges = [Range(0, n)]
    heapq.heapify(ranges)

    iteration = 1
    while len(ranges) > 0:
        r = heapq.heappop(ranges)
        i = (r.left + r.right - 1) // 2
        a[i] = iteration
        if i > r.left:
            heapq.heappush(ranges, Range(r.left,i))
        if r.right > (i+1):
            heapq.heappush(ranges, Range(i+1, r.right))
        iteration += 1

    return a


def main():
    for _ in range(ri()):
        n = ri()
        print(ia_to_s(solve(n)))


if __name__ == '__main__':
    main()