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 PYTHON):

Solution:
import sys
if sys.subversion[0] == "PyPy":
    import io, atexit
    sys.stdout = io.BytesIO()
    atexit.register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
    
    sys.stdin = io.BytesIO(sys.stdin.read())
    input = lambda: sys.stdin.readline().rstrip()
 
RS = raw_input
RI = lambda x=int: map(x,RS().split())
RN = lambda x=int: x(RS())
''' ...................................................................... '''
import heapq as hp

class Pair:
    def __init__(self,l,h):
        self.low = l
        self.high = h

    def getSize(self):
        return self.high-self.low

    def __cmp__(self,other):
        s1 = self.getSize()
        s2 = other.getSize()
        if s1 == s2:
            return self.low > other.low
        return s1<s2
        
        

for _ in xrange(RN()):
    n = RN()
    arr = [0]*(n+1)

    pair = Pair(1,n)
    PQ = [pair]

    for i in xrange(1,n+1):
        pair = hp.heappop(PQ)
        mid = (pair.low+pair.high)/2
        arr[mid] = i

        if pair.low !=mid:
            hp.heappush(PQ,Pair(pair.low,mid-1))
        if pair.high != mid:
            hp.heappush(PQ,Pair(mid+1,pair.high))

    print ' '.join(map(str,arr[1:]))