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,sys
# # import sys

# for strings
# input = sys.stdin.buffer.readline
# import os
# import sys
# from atexit import register
# from io import BytesIO

# sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
# sys.stdout = BytesIO()
# register(lambda: os.write(1, sys.stdout.getvalue()))

# input = lambda: sys.stdin.readline().rstrip('\r\n')

import sys
# import sys
# import heapq
# from atexit import register
# from io import BytesIO

# sys.stdout = BytesIO()
# register(lambda: os.write(1, sys.stdout.getvalue()))

input = sys.stdin.buffer.readline


import heapq
def solution():
    for _ in range(int(input())):
        n=int(input())
        a=[0]*(n)
        h=[(-n,(0,n-1))]
        i=1
        while i<=n:
            len, (l,r)=heapq.heappop(h)
            mid=(l+r)//2
            a[mid]=i
            heapq.heappush(h,(-(mid-l),(l,mid-1)))
            heapq.heappush(h,(-(r-mid),(mid+1,r)))
            i+=1
        print(*a)

solution()