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:
import heapq

T = int(input())
for t in range(T):
    # print("----")
    n = int(input())
    a = [0 for i in range(n)]
    interval_length = n
    prev_pos = [(-n, 0, n - 1)]
    heapq.heapify(prev_pos)
    # print(prev_pos)
    nb = 1
    while True:
        try:
            interval_length, l, r = heapq.heappop(prev_pos)
        except Exception:
            break
        interval_length = -interval_length
        # print("interval_length, l, r", interval_length, l, r)

        if l == r:
            a[l] = nb
            nb += 1
        else:
            middle = (l + r) // 2
            a[middle] = nb
            nb += 1
            interval_length = interval_length // 2
            # print("l, middle, r", l, middle, r)
            if middle - 1 >= l:
                new_l = middle - l
                heapq.heappush(prev_pos, (-new_l, l, middle - 1))
            if middle + 1 <= r:
                new_l = r - middle
                heapq.heappush(prev_pos, (-new_l, middle + 1, r))
    print(" ".join((map(str, a))))


# T = int(input())
# for t in range(T):
#     print("----")
#     n = int(input())
#     if n == 1:
#         a = [1]
#     else:
#         a = [0 for i in range(n)]
#         indexes_used = [False for i in range(n)]
#         pos = [(n - 1) // 2]
#         print("n, pos", n, pos)
#         nb = 1
#         interval_length = (n) // 2
#         print("interval_length : ", interval_length)
#         order = -1 if (n // 2) == 0 else 1
#         while interval_length >= 1:
#             # print("interval", interval_length)
#             interval_length = interval_length // 2
#             new_indexes = []
#             while pos:
#                 idx = pos.pop(0)
#                 # print(idx)
#                 a[idx] = nb
#                 nb += 1
#                 new_indexes.append(idx + order * interval_length)
#                 new_indexes.append(idx + order * interval_length)
#                 order = -1 if order == 1 else 1
#             pos += new_indexes

#     print(" ".join((map(str, a))))


# T = int(input())
# for t in range(T):
#     print("----")
#     n = int(input())
#     a = [0 for i in range(n)]
#     interval_length = (n + 1) // 2
#     prev_positions = [interval_length - 1]
#     found = 1
#     level = 1
#     to_place = 2 ** level
#     last_placed = 1
#     while found <= n:
#         maxi = min(n, last_placed + to_place)
#         for i in range(last_placed + 1, maxi):

#             found += 1
#             last_placed = i

#         level += 1
#         to_place = 2 ** level


# # T = int(input())
# # for t in range(T):
# #     print("----")
# #     n = int(input())
# #     if n == 1:
# #         a = [1]
# #     else:
# #         a = [0 for i in range(n)]
# #         indexes_used = [False for i in range(n)]
# #         pos = [(n - 1) // 2]
# #         print("n, pos", n, pos)
# #         nb = 1
# #         interval_length = (n) // 2
# #         print("interval_length : ", interval_length)
# #         order = -1 if (n // 2) == 0 else 1
# #         while interval_length >= 1:
# #             # print("interval", interval_length)
# #             interval_length = interval_length // 2
# #             new_indexes = []
# #             while pos:
# #                 idx = pos.pop(0)
# #                 # print(idx)
# #                 a[idx] = nb
# #                 nb += 1
# #                 new_indexes.append(idx + order * interval_length)
# #                 new_indexes.append(idx + order * interval_length)
# #                 order = -1 if order == 1 else 1
# #             pos += new_indexes

# #     print(" ".join((map(str, a))))