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:
# cook your dish here
# import sys
# sys.stdin = open('input.txt', 'r') 
# sys.stdout = open('output.txt', 'w')

import math
import collections
from sys import stdin,stdout,setrecursionlimit
import bisect as bs
T = int(stdin.readline())
setrecursionlimit(2**20)
def solve(l,r):
    global i,d
    if(l > r):
        return
    if((r-l+1) not in d):
        d[r-l+1] = []
    d[r-l+1].append(l)
    if(l == r):
        a[l] = i
        i+=1
        return
    if((r-l+1) % 2 == 1):
        a[(l+r)//2] = i
        s = (l+r)//2
    else:
        a[(l+r-1)//2] = i
        s = (l+r)//2
    i += 1
    solve(l, s-1)
    solve(s+1, r )
 
 
for _ in range(T):
    n = int(stdin.readline())
    # n,k = list(map(int,stdin.readline().split()))
    # a = list(map(int,stdin.readline().split()))
    # b = list(map(int,stdin.readline().split()))
    # s = stdin.readline().strip('\n')
    i = 1
    a = [0]*(n+1)
    d = {}
    solve(1,n)
    i = 1
    for s in range(n,0,-1):
        if(s not in d):
            continue
        ls = d[s]
        ls.sort()
        for l in ls:
            r = l+s-1
            if((r-l+1) % 2 == 1):
                a[(l+r)//2] = i
            else:
                a[(l+r-1)//2] = i
            i +=1
    for h in a[1:]:
        print(h,end= ' ')
    print('')