Problem:
Score : 300 points

Problem Statement

Given is a string S. Let T be the concatenation of K copies of S.
We can repeatedly perform the following operation: choose a character in T and replace it with a different character.
Find the minimum number of operations required to satisfy the following condition: any two adjacent characters in T are different.

Constraints

1 \leq |S| \leq 100

S consists of lowercase English letters.

1 \leq K \leq 10^9

K is an integer.

Input

Input is given from Standard Input in the following format:

S
K

Output

Print the minimum number of operations required.

Sample Input 1

issii
2

Sample Output 1

4

T is issiiissii. For example, we can rewrite it into ispiqisyhi, and now any two adjacent characters are different.

Sample Input 2

qq
81

Sample Output 2

81

Sample Input 3

cooooooooonteeeeeeeeeest
999993333

Sample Output 3

8999939997

Solution in OCaml (4.02.3) (judge verdict: Accepted)

let split_string ?(pattern="") = Str.split @@ Str.regexp pattern
let s = read_line () |> split_string |> Array.of_list
let k = Scanf.sscanf (read_line ()) "%d" @@ fun k -> k

let s_len = Array.length s

let f () = 
  let cnt = Array.of_list @@ fst @@ Array.fold_left (fun (acc, before) ch ->
    match acc with
      | [] -> ([1], ch)
      | x :: xs -> ((if before = ch then ((x + 1) :: xs) else 1 :: acc), ch)
  ) ([], "") s in
  if Array.length cnt = 1 then s_len * k / 2
  else
  (cnt |> Array.map (fun i -> i / 2) |> Array.fold_left (+) 0) * k -
    if s.(0) = s.(s_len - 1) then let a, b = cnt.(0), cnt.(Array.length cnt - 1) in (a / 2 + b / 2 - (a + b) / 2) * (k - 1)
    else 0

let () =
  Printf.printf "%d\n" @@ match s_len with
    | 1 -> k / 2
    | 2 -> if s.(0) = s.(1) then k else 0
    | _ -> f ()