A median of an array of integers of length $n$ is the number standing on the $\lceil {\frac{n}{2}} \rceil$ (rounding up) position in the non-decreasing ordering of its elements. Positions are numbered starting with $1$. For example, a median of the array $[2, 6, 4, 1, 3, 5]$ is equal to $3$. There exist some other definitions of the median, but in this problem, we will use the described one.

Given two integers $n$ and $k$ and non-decreasing array of $nk$ integers. Divide all numbers into $k$ arrays of size $n$, such that each number belongs to exactly one array.

You want the sum of medians of all $k$ arrays to be the maximum possible. Find this maximum possible sum.

Input

The first line contains a single integer $t$ ($1 \leq t \leq 100$) — the number of test cases. The next $2t$ lines contain descriptions of test cases.

The first line of the description of each test case contains two integers $n$, $k$ ($1 \leq n, k \leq 1000$).

The second line of the description of each test case contains $nk$ integers $a_1, a_2, \ldots, a_{nk}$ ($0 \leq a_i \leq 10^9$) — given array. It is guaranteed that the array is non-decreasing: $a_1 \leq a_2 \leq \ldots \leq a_{nk}$.

It is guaranteed that the sum of $nk$ for all test cases does not exceed $2 \cdot 10^5$.

Output

For each test case print a single integer — the maximum possible sum of medians of all $k$ arrays.

Note

The examples of possible divisions into arrays for all test cases of the first test:

Test case $1$: $[0, 24], [34, 58], [62, 64], [69, 78]$. The medians are $0, 34, 62, 69$. Their sum is $165$.

Test case $2$: $[27, 61], [81, 91]$. The medians are $27, 81$. Their sum is $108$.

Test case $3$: $[2, 91, 92, 95], [4, 36, 53, 82], [16, 18, 21, 27]$. The medians are $91, 36, 18$. Their sum is $145$.

Test case $4$: $[3, 33, 35], [11, 94, 99], [12, 38, 67], [22, 69, 71]$. The medians are $33, 94, 38, 69$. Their sum is $234$.

Test case $5$: $[11, 41]$. The median is $11$. The sum of the only median is $11$.

Test case $6$: $[1, 1, 1], [1, 1, 1], [1, 1, 1]$. The medians are $1, 1, 1$. Their sum is $3$.