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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.io.PrintWriter
val pw = PrintWriter(System.out)

fun main(args : Array<String>) {
    func()
    pw.flush()
}

fun func() {
    val s = next()
    val k = nextLong()

    if (s.length == 1) {
        println(k / 2L)
        return
    }

    var changed = false
    var count = 0L

    for (i in 1 until s.length) {
        if (changed) {
            changed = false
            continue
        }
        if (s[i] == s[i-1]) {
            count++
            changed = true
        }
    }

    if (k == 1L) {
        println(count)
        return
    }

    // abbc
    if (s.first() != s.last()) {
        println(count * k)
        return
    }

    val ss = s + s
    changed = false
    var dCount = 0L
    for (i in 1 until ss.length) {
        if (changed) {
            changed = false
            continue
        }
        if (ss[i] == ss[i-1]) {
            dCount++
            changed = true
        }
    }


    // aaa
    if (s.first() == s.last() && s.last() == s[s.length - 2]) {
        if (k % 2L == 0L) {
            println(dCount * (k / 2L))
        } else {
            println(dCount * (k / 2L) + count)
        }
        return
    }

    // abba
    if (s.first() == s.last() && s.last() != s[s.length - 2]) {
        dCount++
        if (k % 2L == 0L) {
            println(dCount * (k / 2L) - 1)
        } else {
            println(dCount * (k / 2L) + count - 1)
        }
        return
    }
}

fun next() = readLine()!!
fun nextLong() = next().toLong()

fun println(value : Any) {
    pw.println(value)
}