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 Fortran (gfortran v4.8.4) (judge verdict: Accepted)

program main
implicit none
integer:: i,ind,oc
integer(8):: k, ans
character(101):: rs
character(:), allocatable:: s
integer, allocatable:: counter(:)
character(1):: cfirst, cend, currentchar, os

read*, rs,k
allocate(character(len_trim(rs)):: s)
s = trim(rs)
! print*, len(s)
allocate(counter(len(s)))
cfirst = s(1:1)
cend = s(len(s):len(s))
ind = 0
counter(1:len(s)) = 0
currentchar = '0'
if (len(s) == 1)then
    print*, k/2
    stop
end if


do i=2, len(s)
    os = s(i-1:i-1)
    if (.not. os == s(i:i)) exit
    if (i==len(s))then
        if (mod(k*len(s),2) == 0)then
            print*, (len(s)*k)/2
        else
            print*, (len(s)*k)/2
        end if
        stop
    endif
    
end do





do i=1, len(s)
    if(currentchar == s(i:i))then
        ! print*, '#1-1'
        counter(ind) = counter(ind) + 1
    else
        ! print*, '#1-2'
        currentchar = s(i:i)
        ind = ind + 1
        counter(ind) = counter(ind) + 1
    end if
    ! print*, counter
end do

ans = sum(counter(1:len(s))/2)
! print*, counter 
if (cfirst == cend)then

    oc = counter(1)
    counter(ind) = counter(ind) + counter(1)
    counter(1) = counter(1) - oc
end if

ans = ans + sum(counter(1:len(s))/2)*(k-1)
! print*, counter 

print*, ans
end program