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 collections
import heapq


# Got Hacked
# def solve(n):
#     arr = [0 for _ in range(n+1)]
#     pq = []
#     heapq.heappush(pq, (-(n), 1, n))

#     for i in range(1, n+1):
#         maxLen, l, r = heapq.heappop(pq)

#         if r-l+1 & 1:
#             mid = (l+r)//2
#         else:
#             mid = (l+r-1)//2

#         arr[mid] = i

#         newLOne, newROne = l, mid - 1
#         newLTwo, newRTwo = mid + 1, r

#         heapq.heappush(pq, (-(newROne - newLOne + 1), newLOne, newROne))
#         heapq.heappush(pq, (-(newRTwo - newLTwo + 1), newLTwo, newRTwo))

#     return " ".join(map(str, arr[1:]))

def solve(n):
    arr = [0 for _ in range(n+1)]
    pq = []
    heapq.heappush(pq, (-(n), 1, n))

    for i in range(1, n+1):
        maxLen, l, r = heapq.heappop(pq)

        if r-l+1 & 1:
            mid = (l+r)//2
        else:
            mid = (l+r-1)//2

        arr[mid] = i

        newLOne, newROne = l, mid - 1
        newLTwo, newRTwo = mid + 1, r

        if (newROne - newLOne + 1) > 0:
            heapq.heappush(pq, (-(newROne - newLOne + 1), newLOne, newROne))

        if (newRTwo - newLTwo + 1) > 0:
            heapq.heappush(pq, (-(newRTwo - newLTwo + 1), newLTwo, newRTwo))

    return " ".join(map(str, arr[1:]))


t = int(input().strip())

for i in range(t):
    n = int(input().strip())

    print(solve(n))