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 defaultdict as dc
from collections import Counter 
from heapq import *
import math
from bisect import bisect_left,bisect
#bisect gives x and p[x] is element greater than it and out of bound for last one 
#p[x-1] gives equal or smaller and no error for any element.
import sys
from collections import deque as dq
from heapq import heapify,heappush,heappop
mod=10**9 +7
def sinp():
    p=str(input())
    return p
def seive(n):
    p=[0]*(n+1)
    for i in range(2,n):
        if p[i]==0:
            for j in range(2*i,n+1,i):
                p[i]=1
    return p
def inp(): 
    p=int(input())
    return p
def line():
    p=list(map(int,input().split()))
    return p
def read_mat():
    n=inp()
    a=[]
    for i in range(n):
        a.append(line())
    return a
def digit(n):
    s=str(n)
    p=0
    for i in s:
        p+=(int(i))**2
    return p

def solve(n):
    z=1
    a=[0]*n
    q=[]
    q.append([-n+1,0])
    heapify(q)
    while(len(q)):
        p,l=heappop(q)
        r=l-p
        mid=(l+r)//2
        a[mid]=z
        z+=1
        if l<=mid-1:
            heappush(q,[l-mid+1,l])
        if r>=mid+1:
            heappush(q,[mid+1-r,mid+1])
    return a
for test in range(inp()):
    n=inp()
    l=solve(n)
    print(*l)