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 collections import Counter
from collections import defaultdict
from collections import deque
import math
import heapq
import sys
input = sys.stdin.readline
import bisect
rs = lambda: input().strip()
ri = lambda: int(input())
rl = lambda: list(map(int, input().split()))
rls= lambda: list(map(str, input().split()))

def res(a,b):
    if((a+b)%2==1):
        return (a+b-1)//2
    else:
        return (a+b)//2

t=int(input())
for _ in range(0,t):
    n=int(input())
    q=[]
    ans=[0]*(n+1)
    heapq.heapify(q)
    x=res(1,n)
    heapq.heappush(q,[-(n-1),1,n])
    c=1
    while(len(q)!=0):
       # print(q)
        a=heapq.heappop(q)

        x=res(a[1],a[2])
        ans[x]=c
       # print("ans", a, x, c)
        c=c+1
        if(a[1]!=a[2] and a[1]<=x-1):
            heapq.heappush(q,[-(x-1-a[1]),a[1],x-1])
        if(a[1]!=a[2] and a[2]>=x+1):
            heapq.heappush(q,[-(a[2]-x-1),x+1,a[2]])
    print(*ans[1:])