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:
t = int(input())
for i in range(t):
    n = int(input())
    dictionary = {}


    def search_in_depth(l, r):
        # print(l, r)
        if l <= r:
            if l == r:
                if 1 not in dictionary:
                    dictionary[1] = [[l]]
                else:
                    dictionary[1] += [[l]]
                return
            else:
                if r - l + 1 not in dictionary:
                    dictionary[r - l + 1] = [[l, r]]
                else:
                    dictionary[r - l + 1] += [[l, r]]
                if (l + r) % 2 == 1:
                    search_in_depth(l, (l + r - 1) // 2 - 1)
                    search_in_depth((l + r - 1) // 2 + 1, r)
                else:
                    search_in_depth(l, (l + r) // 2 - 1)
                    search_in_depth((l + r) // 2 + 1, r)


    search_in_depth(1, n)
    sorted_dictionary = {i: dictionary[i] for i in sorted(dictionary, reverse=True)}
    # print(dictionary)
    # print(sorted_dictionary)
    indices = []
    for key in sorted_dictionary:
        for value in sorted_dictionary[key]:
            # print(value)
            if (value[0] + value[-1]) % 2 == 1:
                indices += [(value[0] + value[-1] - 1) // 2 - 1]
            else:
                indices += [(value[0] + value[-1]) // 2 - 1]
    # print(indices)
    answer = [0] * n
    for counter, ind in enumerate(indices, 1):
        answer[ind] = counter
    print(*answer)