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.
'''
======================================
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) —

(pi OR pi+1 OR … OR pj−1 OR pj)≥j−i+1, where OR denotes the bitwise OR operation.
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.
======================================
'''

import itertools

def generateArray(len):
    lst = []
    for i in range(0,len):
        lst.append(i+1)
    return lst

def checkPermutations(lst):
    lest = []
    for perm in lst:
        valids = 0
        if isinstance(perm, int):
            perm = [perm]
        subsLst = []
        subs = printSubArrays(perm, 0, 0, subsLst)
        for sub in subs:
            buffer = sub[0]
            for i in range(0,len(sub)):
                try:
                    buffer = sub[i] | sub[i+1]
                except:
                    pass
            if buffer >= len(sub):
                valids += 1
        if valids == len(subs):
            lest.append(perm)
            return lest

def createPermutations(lst):
    return list(itertools.permutations(lst))


def printSubArrays(arr, start, end, lst):
    if end == len(arr):
        return lst
    elif start > end:
        return printSubArrays(arr, 0, end + 1, lst)
    else:
        lst.append(arr[start:end + 1])
        return printSubArrays(arr, start + 1, end, lst)


tests = int(input())
for i in range(0,tests):
    length = int(input())
    #arr = generateArray(length)
    #perms = createPermutations(arr)
    #answer = checkPermutations(perms)[0]
    answer = generateArray(length)
    strng = " ".join(map(str, answer))
    print(strng)
