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:
import heapq
t=int(input())
for _ in range(t):
    n=int(input())
    d={}
    d[n] = []
    heapq.heapify(d[n])
    heapq.heappush(d[n],1)
    l=[0]*(n+1)
    m=1
    c=n
    p=[]
    heapq.heapify(p)
    while True:
        while d[c]:
            x=heapq.heappop(d[c])
            y = x+n-1
            if c%2:
                e = x+c//2
                l[e]=m
                m+=1
                try:
                    heapq.heappush(d[c//2],x)
                    heapq.heappush(d[c//2],e+1)
                except:
                    d[c//2]=[]
                    heapq.heapify(d[c//2])
                    heapq.heappush(d[c//2],x)
                    heapq.heappush(d[c//2],e+1)
                if c//2>=1:
                    heapq.heappush(p,-1*(c//2))
                    
            else:
                e=x+c//2-1
                l[e]=m
                m+=1
                try:
                    heapq.heappush(d[c//2-1],x)
                except:
                    d[c//2-1]=[]
                    heapq.heapify(d[c//2-1])
                    heapq.heappush(d[c//2-1],x)
                try:
                    heapq.heappush(d[c//2],e+1)
                except:
                    d[c//2]=[]
                    heapq.heapify(d[c//2])
                    heapq.heappush(d[c//2],e+1)
                if c//2>=1:
                    heapq.heappush(p,-1*(c//2))
                if c//2-1>0:
                    heapq.heappush(p,-1*(c//2-1))
                

        if p:
            c=heapq.heappop(p)
            c=c*-1
        else:
            break

    del(l[0])        
    print(*l)