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code_segments/segment_381.txt

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+You are given an array $a = [1, 2, \ldots, n]$, where $n$ is odd, and an integer $k$.
+
+Your task is to choose an odd positive integer $m$ and to split $a$ into $m$ subarrays$^{\dagger}$ $b_1, b_2, \ldots, b_m$ such that:
+
+  * Each element of the array $a$ belongs to exactly one subarray.    * For all $1 \le i \le m$, $|b_i|$ is odd, i.e., the length of each subarray is odd.    * $\operatorname{median}([\operatorname{median}(b_1), \operatorname{median}(b_2), \ldots, \operatorname{median}(b_m)]) = k$, i.e., the median$^{\ddagger}$ of the array of medians of all subarrays must equal $k$. $\operatorname{median}(c)$ denotes the median of the array $c$. 
+
+$^{\dagger}$A subarray of the array $a$ of length $n$ is the array $[a_l, a_{l + 1}, \ldots, a_r]$ for some integers $1 \le l \le r \le n$.
+
+$^{\ddagger}$A median of the array of odd length is the middle element after the array is sorted in non-decreasing order. For example: $\operatorname{median}([1,2,5,4,3]) = 3$, $\operatorname{median}([3,2,1]) = 2$, $\operatorname{median}([2,1,2,1,2,2,2]) = 2$.
+
+Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. The description of the test cases follows.
+
+The first line of each test case contains two integers $n$ and $k$ ($1 \le k \le n < 2 \cdot 10^5$, $n$ is odd) — the length of array $a$ and the desired median of the array of medians of all subarrays.
+
+It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.
+
+For each test case:
+
+  * If there is no suitable partition, output $-1$ in a single line.    * Otherwise, in the first line, output an odd integer $m$ ($1 \le m \le n$), and in the second line, output $m$ distinct integers $p_1, p_2 , p_3 , \ldots, p_m$ ($1 = p_1 < p_2 < p_3 < \ldots < p_m \le n$) — denoting the left borders of each subarray. 
+
+In detail, for a valid answer $[p_1, p_2, \ldots, p_m]$:
+
+  * $b_1 = \left[ a_{p_1}, a_{p_1 + 1}, \ldots, a_{p_2 - 1} \right]$    * $b_2 = \left[ a_{p_2}, a_{p