CF2001B Generate Permutation

There is an integer sequence $ a $ of length $ n $ , where each element is initially $ -1 $ . Misuki has two typewriters where the first one writes letters from left to right, with a pointer initially pointing to $ 1 $ , and another writes letters from right to left with a pointer initially pointing to $ n $ . Misuki would choose one of the typewriters and use it to perform the following operations until $ a $ becomes a permutation of $ [1, 2, \ldots, n] $ - write number: write the minimum positive integer that isn't present in the array $ a $ to the element $ a_i $ , $ i $ is the position where the pointer points at. Such operation can be performed only when $ a_i = -1 $ . - carriage return: return the pointer to its initial position (i.e. $ 1 $ for the first typewriter, $ n $ for the second) - move pointer: move the pointer to the next position, let $ i $ be the position the pointer points at before this operation, if Misuki is using the first typewriter, $ i := i + 1 $ would happen, and $ i := i - 1 $ otherwise. Such operation can be performed only if after the operation, $ 1 \le i \le n $ holds. Your task is to construct any permutation $ p $ of length $ n $ , such that the minimum number of carriage return operations needed to make $ a = p $ is the same no matter which typewriter Misuki is using.

Each test contains multiple test cases. The first line of input contains a single integer $ t $ ( $ 1 \le t \le 500 $ ) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $ n $ ( $ 1 \le n \le 2 \cdot 10^5 $ ) — the length of the permutation. It is guaranteed that the sum of $ n $ over all test cases does not exceed $ 2 \cdot 10^5 $ .

For each test case, output a line of $ n $ integers, representing the permutation $ p $ of length $ n $ such that the minimum number of carriage return operations needed to make $ a = p $ is the same no matter which typewriter Misuki is using, or $ -1 $ if it is impossible to do so. If there are multiple valid permutations, you can output any of them.

In the first testcase, it's possible to make $ a = p = [1] $ using $ 0 $ carriage return operations. In the second testcase, it is possible to make $ a = p = [1, 2] $ with the minimal number of carriage returns as follows: If Misuki is using the first typewriter: - Write number: write $ 1 $ to $ a_1 $ , $ a $ becomes $ [1, -1] $ - Move pointer: move the pointer to the next position. (i.e. $ 2 $ ) - Write number: write $ 2 $ to $ a_2 $ , $ a $ becomes $ [1, 2] $ If Misuki is using the second typewriter: - Move pointer: move the pointer to the next position. (i.e. $ 1 $ ) - Write number: write $ 1 $ to $ a_1 $ , $ a $ becomes $ [1, -1] $ - Carriage return: return the pointer to $ 2 $ . - Write number: write $ 2 $ to $ a_2 $ , $ a $ becomes $ [1, 2] $ It can be proven that the minimum number of carriage returns needed to transform $ a $ into $ p $ when using the first typewriter is $ 0 $ and it is $ 1 $ when using the second one, so this permutation is not valid. Similarly, $ p = [2, 1] $ is also not valid, so there is no solution for $ n = 2 $ . In the third testcase, it is possibile to make $ a = p = [3, 1, 2] $ with $ 1 $ carriage return with both the first and the second typewriter. It can be proven that, for both typewriters, it is impossible to write $ p $ with $ 0 $ carriage returns. With the first typewriter it is possible to: - Move pointer: move the pointer to the next position. (i.e. $ 2 $ ) - Write number: write $ 1 $ to $ a_2 $ , $ a $ becomes $ [-1, 1, -1] $ - Move pointer: move the pointer to the next position. (i.e. $ 3 $ ) - Write number: write $ 2 $ to $ a_3 $ , $ a $ becomes $ [-1, 1, 2] $ - Carriage return: return the pointer to $ 1 $ . - Write number: write $ 3 $ to $ a_1 $ , $ a $ becomes $ [3,1,2] $ With the second typewriter it is possible to: - Move pointer: move the pointer to the next position. (i.e. $ 2 $ ) - Write number: write $ 1 $ to $ a_2 $ , $ a $ becomes $ [-1, 1, -1] $ - Carriage return: return the pointer to $ 3 $ . - Write number: write $ 2 $ to $ a_3 $ , $ a $ becomes $ [-1, 1, 2] $ - Move pointer: move the pointer to the next position. (i.e. $ 2 $ ) - Move pointer: move the pointer to the next position. (i.e. $ 1 $ ) - Write number: write $ 3 $ to $ a_1 $ , $ a $ becomes $ [3, 1, 2] $