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
for _ in range(int(input())):
    n = int(input())
    a = [0]*n
    d = collections.deque()
    ans = [[1, n]]
    d.append([1, n])
    for i in range(1, n+1):
        s = d.popleft()
        r = s[1]
        l = s[0]
        # print(a, l, r)
        if (r - l + 1) % 2 == 1:
            a[(l + r)// 2 - 1] = i
            if r - l == 1:
                d.append([(l + r)// 2 + 1, r])
                ans.append([(l + r)// 2 + 1, r])
            elif r - l >= 2:
                d.append([l, (l + r)// 2 - 1])
                ans.append([l, (l + r)// 2 - 1])
                d.append([(l + r)// 2 + 1, r])
                ans.append([(l + r)// 2 + 1, r])
        else:
            a[(l + r - 1) // 2 - 1] = i
            if r - l == 1:
                d.append([(l + r)// 2 + 1, r])
                ans.append([(l + r)// 2 + 1, r])
            elif r - l >= 2:
                d.append([(l + r) // 2 + 1, r])
                ans.append([(l + r) // 2 + 1, r])
                d.append([l, (l + r) // 2 - 1])
                ans.append([l, (l + r) // 2 - 1])
    ans.sort(key = lambda x: (-(x[1] - x[0]), x[0]))
    k = 1
    for i in ans:
        r = i[1]
        l = i[0]
        if (r - l + 1) % 2 == 1:
            a[(l + r)// 2 - 1] = k
        else:
            a[(l + r - 1) // 2 - 1] = k
        k += 1

    # print(ans)
    print(*a)