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 math


def task(n):
    table = {}
    for i in range(0, n):
        table[i] = []

    table[n] = [[1, n, 1]]
    array = [0] * n
    i = 0

    for size in reversed(range(1, n + 1)):
        if len(table[size]) == 0:
            continue

        intervals = sorted(table[size], key=lambda x: (x[0]))
        for interval in intervals:
            i += 1
            l, r = interval[0], interval[1]

            if (r - l + 1) % 2 == 0:
                index = (l + r - 1) // 2
            else:
                index = (l + r) // 2

            array[index - 1] = i

            if l <= index - 1:
                intervalsize = index - 1 - l + 1
                table[intervalsize].append([l, index - 1])

            if index + 1 <= r:
                intervalsize = r - index - 1 + 1
                table[intervalsize].append([index + 1, r])

            # print(l, r)

    print(" ".join(map(str, array)))


t = int(input())
for i in range(0, t):
    n = int(input())
    task(n)