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 sys import stdout, stdin
_input, _print = stdin.buffer.readline, stdout.write
_int, _range, _len = int, range, len
def heap_push(h, v):
    i = _len(h)
    h.append(v)
    while i-1 // 2 > 0 and v[0] >= h[(i-1)//2][0]:
            if h[(i-1)//2][0] == v[0]:
                if v[1] < h[(i-1)//2][1]:
                    h[(i - 1) // 2], h[i] = h[i], h[(i - 1) // 2]
                else:
                    break
            else:
                h[(i - 1) // 2], h[i] = h[i], h[(i - 1) // 2]
            i = (i - 1) // 2
def heap_pop(h):
    l = _len(h) - 1
    h[0], h[l] = h[l], h[0]
    i = 0
    while i * 2 + 2 < l:
        if h[i*2+2][0] > h[i*2+1][0]:
            temp = i*2+2
        else:
            if h[i*2+2][0] == h[i*2+1][0]:
                if h[i*2+2][1] < h[i*2+1][1]:
                    temp = i*2+2
                else:
                    temp = i*2+1
            else:
                temp = i*2+1
        if h[i][0] < h[temp][0]:
            h[i], h[temp] = h[temp], h[i]
        elif h[i][0] == h[temp][0] and h[i][1] > h[temp][1]:
            h[i], h[temp] = h[temp], h[i]
        else:
            break
        i = temp
    else:
        if i*2+1 < l:
            if h[i][0] < h[i * 2 + 1][0]:
                h[i], h[i * 2 + 1] = h[i * 2 + 1], h[i]
            else:
                if h[i][0] == h[i * 2 + 1][0] and h[i][1] > h[i * 2 + 1][1]:
                    h[i], h[i * 2 + 1] = h[i * 2 + 1], h[i]
    el = h[-1]
    h.pop()
    return el
def solution():
    for _ in _range(_int(_input())):
        n = _int(_input())
        a = [0]*n
        h = [(n - 1, 1, n)]
        for i in _range(n):
            v, l, r = heap_pop(h)
            if (r - l) % 2 == 0:
                temp = ((r+l)//2)
            else:
                temp = ((r+l-1)//2)
            a[temp - 1] = i + 1
            if temp > l:
                heap_push(h, (temp - 1 - l, l, temp - 1))
            if temp < r:
                heap_push(h, (r - (temp+1), temp+1, r))
        print(*a)
solution()