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:
class Btree:
    l = []
    def shiftup(self, index):
        value = self.l[index]
        while index>0:
            parent = (index-1) >> 1
            parentval = self.l[parent]
            if value < parentval:
                self.l[index] = parentval
                index = parent
                continue
            break
        self.l[index] = value
    def push(self, el):
        self.l.append(el)
        index = len(self.l)-1
        self.shiftup(index)
    def pop(self):
        ans = self.l[0]
        value = self.l.pop()
        if self.l:
            index = 0
            sonl = 1
            totlen = len(self.l)
            while(sonl < totlen):
                sonr = sonl+1
                if sonr < totlen:
                    if self.l[sonl] > self.l[sonr]:
                        sonl = sonr
                self.l[index] = self.l[sonl]
                index = sonl
                sonl = 2*index+1
            self.l[index] = value
            self.shiftup(index)
        return ans
def tuplmk(a,b):
    return (a-1-b, a, b)
def ans(n):
    ans = [0 for i in range(n)]
    h = Btree()
    h.push(tuplmk(1,n))
    for el in range(1,n+1):
        seq = h.pop()
        ind = (seq[1]+seq[2])//2
        ans[ind-1] = el
        h.push(tuplmk(seq[1], ind-1))
        h.push(tuplmk(ind+1, seq[2]))
    return " ".join([str(i) for i in ans])
t = int(input())
tlist = [0 for i in range(t)]
d = {}
for pset in range(t):
    n = int(input())
    tlist[pset] = n
    if n in d:
        print(d[n])
    else:
        answ = ans(n)
        d[n] = answ
        print(answ)