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 input1(type = int):
  return type(input())

def input2(type = int):
  [a, b] = list(map(type, input().split()))
  return a, b

def input3(type = int):
  [a, b, c] = list(map(type, input().split()))
  return a, b, c

def input_array(type = int):
  return list(map(type, input().split()))

def input_string():
  s = input()
  return list(s)

# from queue import PriorityQueue

def get_l(sz, m):
  md = sz // 2
  if sz % 2 == 0:
    md -= 1
  return m - md

def get_r(sz, m):
  return m + sz//2

def mid_point(l, r):
  length = r - l + 1
  m = length // 2
  if length % 2 == 0:
    m -= 1
  return l + m

def break_here(l, r):
  m = mid_point(l, r)
  sz1 = (m-1) - l + 1
  sz2 = r - (m+1) + 1
  return [(-sz1, mid_point(l, m-1)), (-sz2, mid_point(m+1, r))]

import heapq as pq


def main():
  t = input1()
  for ci in range(t):
    n = input1()

    # pq = heapq()
    h = []
    pq.heappush(h, (-n, mid_point(0, n-1)))

    ind = 1
    arr = [0 for _ in range(n)]
    while len(h):
      now = pq.heappop(h)

      arr[now[1]] = str(ind)
      ind += 1

      [a, b] = break_here(get_l(-now[0], now[1]), get_r(-now[0], now[1]))
      # print(a, b)
      if a[0] < 0:
        pq.heappush(h, a)
      if b[0] < 0:
        pq.heappush(h, b)

    print(" ".join(arr))

  return

main()