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 sys
# from math
import bisect
import heapq
# from collections import deque

# from types import GeneratorType
# def bootstrap(func, stack=[]):
#     def wrapped_function(*args, **kwargs):
#         if stack:
#             return func(*args, **kwargs)
#         else:
#             call = func(*args, **kwargs)
#             while True:
#                 if type(call) is GeneratorType:
#                     stack.append(call)
#                     call = next(call)
#                 else:
#                     stack.pop()
#                     if not stack:
#                         break
#                     call = stack[-1].send(call)
#             return call
#     return wrapped_function

Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
 
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10**9+7

for _ in range(int(ri())):
    n = int(ri())
    ans = [0]*n
    h = [(-n,(0,n-1))]
    heapq.heapify(h)
    ite = 1
    while(len(h) > 0):

        top = heapq.heappop(h)
        l,r = top[1]
        length = -1*top[0]
        if l > r :
            continue
        if l == r:
            ans[l] = ite
        else:
            if length %2 == 1:
                mid = (l+r)//2
                ans[mid] = ite
                heapq.heappush(h,(-(length-1)//2, (l, mid-1)))
                heapq.heappush(h,(-(length-1)//2, (mid+1, r)))
            else:
                mid = (l+r-1)//2
                ans[mid]  = ite
                heapq.heappush(h,(-(length//2-1), (l, mid-1)))
                heapq.heappush(h,(-length//2, (mid+1, r)))
        ite+=1

    print(*ans)