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
input = sys.stdin.readline


def inp():
    return (int(input()))


def inlt():
    return (list(map(int, input().split())))


def insr():
    s = input()
    return (list(s[:len(s) - 1]))


def invr():
    return list((map(int, input().split())))


def solve(num):
    res = [0] * num
    count = 1
    q = [(-num, (0, num-1))]
    while q:
        item = heapq.heappop(q)
        # print(item)
        # print(res)
        n = -item[0]
        l = item[1][0]
        r = item[1][1]
        if l == r:
            res[int(l)] = count
            count += 1
            continue
        if r == l + 1:
            res[int(l)] = count
            count += 1
            heapq.heappush(q, (-1, (r, r)))
            continue
        if (l+r) % 2 == 0:
            m = (l+r) / 2
            heapq.heappush(q, (-(m - l), (l, m-1)))
            heapq.heappush(q, (-(r - m), (m+1, r)))
        else:
            m = (l + r - 1) / 2
            heapq.heappush(q, (-(m - l), (l, m - 1)))
            heapq.heappush(q, (-(r - m), (m + 1, r)))
        res[int(m)] = count
        count += 1
    return res


if __name__ == '__main__':
    t = inp()
    for i in range(t):
        print(" ".join(map(str, solve(inp()))))