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 heapq
import sys
input = sys.stdin.readline


for _ in xrange(int(input())):
    n = int(input())
    ans = [0 for _ in xrange(n)]
    zeros = [((-n*10**6+(n-1)//2), (n-1)//2), ]
    cnt = 1
    while cnt <= n:
        val, mid = heapq.heappop(zeros)
        val = -(val - mid ) // 10**6
        
        ans[mid] = str(cnt)
        vala = (val - 1) // 2
        valb = val - vala - 1
        # print(mid, val, vala, valb, ans)
        if vala > 0:
            loc = mid-(vala+2)//2
            # print(loc, vala, valb)
            # print('loc - a', loc, vala)
            if loc >= 0:
                heapq.heappush(zeros, (-vala*10**6 + loc, loc))
        if valb > 0:
            loc = mid + (valb+1) // 2
            # print('loc - b',loc, valb)
            if loc <n:
                heapq.heappush(zeros, (-valb*10**6 + loc, loc))
        
        cnt += 1
    print(' '.join(ans))