A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) — 

  * (p_i  OR  p_{i+1}  OR  …  OR  p_{j-1}  OR  p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)



In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray. 

Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.

The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).

Output

For every test, output any good permutation of length n on a separate line. 

Example

Input


3
1
3
7


Output


1
3 1 2
4 3 5 2 7 1 6

Note

For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below. 

  * 3 OR 1 = 3 ≥ 2 (i = 1,j = 2) 
  * 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3) 
  * 1 OR 2 = 3 ≥ 2 (i = 2,j = 3) 
  * 1 ≥ 1 (i = 2,j = 2) 



Similarly, you can verify that [4,3,5,2,7,1,6] is also good.
import sys

TEST = 0

if TEST == 1:
    input_file = open("inputA.txt", "r")
    input_from_file = input_file.readlines()

def read_int():
    if TEST:
        return int(input_from_file.pop(0))
    else:
        return int(sys.stdin.readline())

def read_ints():
    if TEST:
        return list(map(int, input_from_file.pop(0).rstrip().split()))
    else:
        return list(map(int, sys.stdin.readline().split()))

def read_str_to_list():
    if TEST:
        in1 = input_from_file.pop(0).rstrip()
    else:
        in1 = sys.stdin.readline().rstrip()
    return [x for x in in1]


tests = read_int()

def isSeqOk(lst):
    if len(lst) == 1:
        return True
    if len(lst) == 2:
        return lst[1] | lst[2] >= 2

    for i in range(len(lst)-1):
        for j in range(i+1, len(lst)):
            x = lst[i] | lst[j]
            if j-i > 1:
                x = x | lst[i+1]
            if j-i > 2:
                x = x | lst[j-1]

            if x < (j+1) - (i+1) + 1:
                return False

    return True

ch = isSeqOk([1, 1, 5, 4, 7, 1, 1])
# print(ch)

for test in range(tests):
    n = read_int()
    if n == 1:
        print("1")
        continue

    mid = int((n)/2)
    u, d = mid+1, mid
    ans = []
    k = 0
    while True:
        if u <= n:
            ans.append(u)
            u += 1
            k += 1

        if d > 0:
            ans.append(d)
            d -= 1
            k += 1

        if k == n:
            break


    print(*ans)







