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 LuaJIT (2.0.4) (judge verdict: Accepted)

local s = io.read()
local k = io.read("*n")
local n = #s
s = s .. s
local t = {}
for i = 1, 2 * n do
  t[i] = s:sub(i, i)
end
local spos = 1
for i = 1, n - 1 do
  if t[i] ~= t[i + 1] then
    spos = i + 1
    break
  end
end
if spos == 1 then
  print(math.floor(n * k / 2))
else
  local edgelen = spos - 1
  local epos = n
  local ecnt = 0
  for i = n, 1, -1 do
    if t[1] == t[i] then
      ecnt = ecnt + 1
    else
      epos = i
      break
    end
  end
  edgelen = edgelen + ecnt
  local intcost = 0
  local len = 1
  for i = spos + 1, epos do
    if t[i] == t[i - 1] then
      len = len + 1
      if len % 2 == 0 then
        intcost = intcost + 1
      end
    else
      len = 1
    end
    -- print(i, len)
  end
  local edgecost = math.floor(edgelen / 2)
  -- print(intcost)
  -- print(intcost * k + edgecost * (k - 1))
  -- print(edgecost)
  -- print(intcost * k, edgecost * (k - 1),  math.floor((spos - 1) / 2) , math.floor(ecnt / 2))
  print(intcost * k + edgecost * (k - 1) + math.floor((spos - 1) / 2) + math.floor(ecnt / 2))
end
--  8999939997
-- 11999919996