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 heapq import *
import sys 
input = lambda : sys.stdin.readline().strip()
t = int(input())
while t:
    t-=1
    n = int(input())
    h = []
    ans = [0]*n
    ans[(n-1)//2] = 1
    cur = 2
    l1 = 0
    r1 = ((n-1)//2)-1
    l2 = (n-1)//2 + 1
    r2 = n-1
    if l1<=r1:
        heappush(h,[-(r1-l1+1),l1])
    if l2<=r2:
        heappush(h,[-(r2-l2+1),l2])
    while h:
        # print(h)
        v = heappop(h)
        ans[v[1]+((-v[0])-1)//2] = cur
        cur+=1
        l1 = v[1]
        r1 = v[1]+(-v[0]-1)//2-1
        l2 = v[1]+(-v[0]-1)//2+1
        r2 = v[1]+(-v[0]-1)
        
        if l1<=r1:
            heappush(h,[-(r1-l1+1),l1])
        if l2<=r2:
            heappush(h,[-(r2-l2+1),l2])
    print(' '.join(map(str,ans)))