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.
__author__ = 'Devesh Bajpai'

'''
https://codeforces.com/problemset/problem/1391/A

Solution: The interesting bit (pun intended) about this problem is that when 2 numbers are bitwise ORed, the bits set
in either of those numbers are set in the result. That means the result number would be at least as much as the 
larger of those 2 numbers. This leads to a postulate that:

a_i OR a_i+1 OR a_i+2 .... a_k >= max(a_i, a_i+1, ... a_k). Hence any ordering of the array 1 to n would satisfy the
condition mentioned in the question. We just return them in their ascending order for ease of implementation. 
'''


def solve(n):

    return " ".join(str(_) for _ in xrange(1, n+1))


if __name__ == "__main__":

    t = int(raw_input())

    results = list()
    for _ in xrange(0, t):
        n = int(raw_input())
        results.append(solve(n))

    for result in results:
        print result
