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
import heapq

def answer(n):
    heap = [] #contain closed tuples [l, r] in the form of (-width, mid, l, r)
    ans = [0 for _ in range(n)]
    l = 0
    r = n-1
    w = -(r-l + 1)
    mid = (r + l) // 2
    heapq.heappush(heap, (w, mid, l, r))
    for ctr in range(1, n+1):
        tup = heapq.heappop(heap)
        l = tup[2]
        r = tup[3]
        mid = tup[1]
        ans[mid] = ctr
        #create left, throw on heap
        ll = l
        lr = mid-1
        lmid = (lr + ll) // 2
        lw = -(lr - ll + 1)
        heapq.heappush(heap, (lw, lmid, ll, lr))
        #create right, throw on heap
        rl = mid+1
        rr = r
        rmid =(rr + rl) // 2
        rw = -(rr - rl + 1)
        heapq.heappush(heap, (rw, rmid, rl, rr))
    
    return ans #list of ints

def main():
    t = int(sys.stdin.readline())
    while t:
        n = int(sys.stdin.readline())
        print(*answer(n))
        t -= 1
    return
main()