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:  Choose the maximum by length subarray (continuous subsegment) consisting only of zeros, among all such segments choose the leftmost one;  Let this segment be $[l; r]$. If $r-l+1$ is odd (not divisible by $2$) then assign (set) $a[\frac{l+r}{2}] := i$ (where $i$ is the number of the current action), otherwise (if $r-l+1$ is even) assign (set) $a[\frac{l+r-1}{2}] := i$. 
Consider the array $a$ of length $5$ (initially $a=[0, 0, 0, 0, 0]$). Then it changes as follows:  Firstly, we choose the segment $[1; 5]$ and assign $a[3] := 1$, so $a$ becomes $[0, 0, 1, 0, 0]$;  then we choose the segment $[1; 2]$ and assign $a[1] := 2$, so $a$ becomes $[2, 0, 1, 0, 0]$;  then we choose the segment $[4; 5]$ and assign $a[4] := 3$, so $a$ becomes $[2, 0, 1, 3, 0]$;  then we choose the segment $[2; 2]$ and assign $a[2] := 4$, so $a$ becomes $[2, 4, 1, 3, 0]$;  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 $t$ ($1 \le t \le 10^4$) — the number of test cases. Then $t$ test cases follow.
The only line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$.
It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----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
import heapq
for _ in range(int(input())):
    n=int(input())
    a=[0]*n
    que=[(-n,0,n-1)]
    heapq.heapify(que)
    i=1
    while que:
        len,l,r=heapq.heappop(que)
        if len%2==1:
            a[(l+r)//2]=i
            if r!=l:
                nl=(-(r-l)//2,l,(l+r)//2-1)
                nr=(-(r-l)//2,(l+r)//2+1,r)
                heapq.heappush(que,nl)
                heapq.heappush(que,nr)
        else:
            a[(l+r-1)//2]=i
            nl=(-(r-l-1)//2,l,(l+r-1)//2-1)
            nr=(-(r-l+1)//2,(l+r-1)//2+1,r)
            if r==l+1:
                heapq.heappush(que,nr)
            else:
                heapq.heappush(que,nr)
                heapq.heappush(que,nl)
        i+=1
    print(*a)
import heapq
for _ in range(int(input())):
    n=int(input())
    a=[0]*n
    que=[(-n,0,n-1)]
    heapq.heapify(que)
    i=1
    while que:
        len,l,r=heapq.heappop(que)
        if len%2==1:
            a[(l+r)//2]=i
            if r!=l:
                nl=(-(r-l)//2,l,(l+r)//2-1)
                nr=(-(r-l)//2,(l+r)//2+1,r)
                heapq.heappush(que,nl)
                heapq.heappush(que,nr)
        else:
            a[(l+r-1)//2]=i
            nl=(-(r-l-1)//2,l,(l+r-1)//2-1)
            nr=(-(r-l+1)//2,(l+r-1)//2+1,r)
            if r==l+1:
                heapq.heappush(que,nr)
            else:
                heapq.heappush(que,nr)
                heapq.heappush(que,nl)
        i+=1
    print(*a)