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: $\operatorname{median}([1,2,5,4,3]) = 3$, $\operatorname{median}([3,2,1]) = 2$, $\operatorname{median}([2,1,2,1,2,2,2]) = 2$.

Input

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 $k$ ($1 \le k \le n \lt 2 \cdot 10^5$, $n$ 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 $2 \cdot 10^5$.

Output

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 $p_1, p_2 , p_3 , \ldots, p_m$ ($1 = p_1 \lt p_2 \lt p_3 \lt \ldots \lt p_m \le n$) — denoting the left borders of each subarray.

In detail, for a valid answer $[p_1, p_2, \ldots, p_m]$:

• $b_1 = \left[ a_{p_1}, a_{p_1 + 1}, \ldots, a_{p_2 - 1} \right]$
• $b_2 = \left[ a_{p_2}, a_{p_2 + 1}, \ldots, a_{p_3 - 1} \right]$
• $\ldots$
• $b_m = \left[ a_{p_m}, a_{p_m + 1}, \ldots, a_n \right]$.

If there are multiple solutions, you can output any of them.

Note

In the first test case, the given partition has $m = 1$ and $b_1 = [1]$. It is obvious that $\operatorname{median}([\operatorname{median}([1])]) = \operatorname{median}([1]) = 1$.

In the second test case, the given partition has $m = 3$ and:

• $b_1 = [1]$
• $b_2 = [2]$
• $b_3 = [3]$

Therefore, $\operatorname{median}([\operatorname{median}([1]), \operatorname{median}([2]), \operatorname{median}([3])]) = \operatorname{median}([1, 2, 3]) = 2$.

In the third test case, there is no valid partition for $k = 3$.

In the fourth test case, the given partition has $m = 5$ and:

• $b_1 = [1, 2, 3]$
• $b_2 = [4, 5, 6]$
• $b_3 = [7, 8, 9]$
• $b_4 = [10, 11, 12]$
• $b_5 = [13, 14, 15]$

Therefore, $\operatorname{median}([\operatorname{median}([1, 2, 3]), \operatorname{median}([4, 5, 6]), \operatorname{median}([7, 8, 9]), \operatorname{median}([10, 11, 12]), \operatorname{median}([13, 14, 15])]) = \operatorname{median}([2, 5, 8, 11, 14]) = 8$.