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

def find_array(n):
    
    lo, hi = 0, n - 1
    heap = [(-(hi - lo + 1), lo, hi)]

    pos = []
    while heap:
        u = heapq.heappop(heap)
        lo, hi = u[1], u[2]
        mid = (lo + hi) >> 1
        pos.append(mid)
        alo, ahi = lo, mid - 1
        blo, bhi = mid + 1, hi
        len_a, len_b = ahi - alo + 1, bhi - blo + 1

        if len_a > 0:
            heapq.heappush(heap, (-len_a, alo, ahi))
        if len_b > 0:
            heapq.heappush(heap, (-len_b, blo, bhi))
    
    res = [0] * len(pos)

    for i in range(len(pos)):
        res[pos[i]] = i + 1
    
    return res

def main():
    t = int(input())

    for i in range(t):
        n = int(input())
        arr = find_array(n)
        for j in range(len(arr)):
            if j == len(arr) - 1:
                print(arr[j])
            else:
                print(arr[j], end=' ')

if __name__ == '__main__':
    main()