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:
from sys import stdin, stdout

int_in = lambda: int(stdin.readline())
arr_in = lambda: [int(x) for x in stdin.readline().split()]
mat_in = lambda rows: [arr_in() for y in range(rows)]
str_in = lambda: stdin.readline().strip()
out = lambda o: stdout.write("{}\n".format(o))
arr_out = lambda o: out(" ".join(map(str, o)))
bool_out = lambda o: out("YES" if o else "NO")

from heapq import heappush, heappop
def solve(n):
    arr = [0]*n
    h = []
    if n > 1:
        heappush(h, (n-1, 0, n-1))
    i = 1
    while len(h) > 0:
        curr_range = heappop(h)[1:]
        curr = sum(curr_range)
        mid = curr // 2
        arr[mid] = i
        i += 1
        left = (curr_range[0], mid-1)
        right = (mid+1, curr_range[1])
        if curr % 2 == 0:
            if left[1] - left[0] > 0:
                heappush(h, (left[0]-left[1], left[0], left[1]))
            if right[1] - right[0] > 0:
                heappush(h, (right[0]-right[1], right[0], right[1]))
        else:
            if right[1] - right[0] > 0:
                heappush(h, (right[0]-right[1], right[0], right[1]))
            if left[1] - left[0] > 0:
                heappush(h, (left[0]-left[1], left[0], left[1]))
    for j in range(n):
        if arr[j] == 0:
            arr[j] = i
            i += 1

    return arr


if __name__ == "__main__":
    for i in range(int_in()):
        n = int_in()
        arr_out(solve(n))