Solve the programming task below in a Python markdown code block.
Read problems statements in Mandarin chinese, Russian and Vietnamese as well. 

You work as an engineer. You were given an empty board with $K$ consecutive cells; at any moment, each cell can display one character.

You want the board to display a string $S$ with length $N > K$. Since the board isn't large enough, you want to display the string in $N-K+1$ steps. In the $i$-th step ($1 ≤ i ≤ N-K+1$), you'll make the board display the characters $S_{i}, S_{i+1}, \dots, S_{i+K-1}$.

The power required to switch the board from step $i$ to step $i+1$ (for $1 ≤ i ≤ N-K$) is equal to the number of characters displayed on the board that have to change between these steps. You should find the total power required for the whole process of displaying a string, i.e. the sum of powers required for switching between all consecutive pairs of steps.

------  Input ------
The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
The first line of each test case contains two space-separated integers $N$ and $K$.
The second line contains a single string $S$ with length $N$.

------  Output ------
For each test case, print a single line containing one integer — the total power required for text switching.

------  Constraints ------
$1 ≤ T ≤ 1,000$
$2 ≤ N ≤ 10^{5}$
$1 ≤ K < N$
each character of $S$ is a lowercase English letter
the sum of $N$ for all test cases does not exceed $10^{6}$

------  Subtasks ------
Subtask #1 (20 points):
$1 ≤ T ≤ 100$
$2 ≤ N ≤ 50$

Subtask #2 (80 points): original constraints

----- Sample Input 1 ------ 
3
6 3
aabbcc
5 2
abccc
4 3
aabb
----- Sample Output 1 ------ 
4
3
1
----- explanation 1 ------ 
Example case 1:
- In step $1$, the board is displaying "aab".
- In step $2$, the board is displaying "abb".
- In step $3$, the board is displaying "bbc".
- In step $4$, the board is displaying "bcc".

The power required for switching from the $1$-st to the $2$-nd step is $1$, because cell $1$ changes from 'a' to 'a' (requiring power $0$), cell $2$ changes from 'a' to 'b' (requiring power $1$) and cell $3$ changes from 'b' to 'b' (requiring power $0$); $0 + 1 + 0 = 1$.

The power required for switching between the $2$-nd and $3$-rd step is $2$ and the power required for switching between the $3$-rd and $4$-th step is $1$.

Therefore, the answer is $1 + 2 + 1 = 4$.