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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String S = sc.next();
		String S_2 = S + S;
		String S_3 = S + S + S;
		int N = Integer.parseInt(sc.next());
		char first = S.charAt(0);
		char before = first;
		long cnt_1 = 0L;
		for (int i=1;i<S.length();i++) {
			if (S.charAt(i)==before) {
				cnt_1++;
				i++;
			}
			if (i<=S.length()-1) {
				before=S.charAt(i);
			}
		}

		long cnt_2 = 0L;
		int flag=0;
		before = first;
		for (int i=1;i<S_2.length();i++) {
			if (S_2.charAt(i)==before) {
				cnt_2++;
				i++;
			}
			if (i<=S_2.length()-1) {
				before=S_2.charAt(i);
			}
			if (i==S_2.length()-1) {
				if (S_2.charAt(i)==first) {
					flag=1;
					if (S.charAt(0)==S.charAt(1) && S_2.charAt(i)==first) {
						flag=0;
					}
				}
			}
		}

		long cnt_3 = 0L;
		before = first;
		for (int i=1;i<S_3.length();i++) {
			if (S_3.charAt(i)==before) {
				cnt_3++;
				i++;
			}
			if (i<=S_3.length()-1) {
				before=S_3.charAt(i);
			}
		}

		long ans = 0L;
		if (N%2==0) {
			// ans = cnt_2*(N/2)+Math.max(flag*(N/2)-1, 0);
			// ans = (cnt_2-cnt_1) * (N-1) + cnt_1;
			ans = (cnt_3-cnt_1) * (N/2-1) + cnt_2;
		} else {
			// ans = cnt_2*(N/2)+cnt_1;
			// if (flag==1) {
				// ans+=flag*(N/2);
			// }
			// ans = (cnt_2-cnt_1) * (N-1) + cnt_1;
			ans = (cnt_3-cnt_1) * (N/2) + cnt_1;
		}
		System.out.println(ans);
		// System.out.println(cnt_1);
		// System.out.println(cnt_2);
		// System.out.println(flag);
	}
}