You are given an integer N.

Find $a permutation P = [P_{1}, P_{2}, \ldots, P_{N}]$ of the integers ${1, 2, \ldots, N$} such that sum of averages of all consecutive triplets is minimized, i.e.

$\sum_{i=1}^{N-2} \frac{P_{i} + P_{i+1} + P_{i+2}}{3}$

is minimized.

If multiple permutations are possible, print any of them.

------ Input Format ------

• The first line of input will contain a single integer T, denoting the number of test cases.
• The first and only line of each test case contains an integer N, the size of the permutation.

------ Output Format ------

For each test case, output on a new line a permutation which satisfies the above conditions.

------ Constraints ------

$1 ≤ T ≤ 1000$
$3 ≤ N ≤ 10^{5}$
- The sum of $N$ over all test cases won't exceed $3\cdot 10^{5}$.

----- Sample Input 1 ------
2
4
3

----- Sample Output 1 ------
3 2 1 4
3 2 1
----- explanation 1 ------
Test case $1$: The sum is $\frac{P_{1} + P_{2} + P_{3}}{3} + \frac{P_{2} + P_{3} + P_{4}}{3} = \frac{3 + 2 + 1}{3} + \frac{2 + 1 + 4}{3} = 6/3 + 7/3 = 4.333\ldots$ Among all possible permutations of $\{1, 2, 3, 4\}$, this is one of the permutations which provides the minimum result.

Test case $2$: The sum is $\frac{3+2+1}{3} = 6/3 = 2$. Every permutation of size $3$ will have this value, hence it is the minimum possible.

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