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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int
main(int argc, char *argv[])
{
    char S[101];
    long long K;
    long long tdup, idup, edup, duptmp;
    int i;

    tdup = 1;
    idup = 0;
    edup = 0;

    scanf("%s", S);
    scanf("%lld", &K);

    for (i=0; i<strlen(S)-1; i++) {
        if (S[i]==S[i+1]) {
            tdup++;
        }
        else
        {
            break;
        }
        
    }
    if(i<strlen(S)-1) {
        duptmp = 1;
        for(; i<strlen(S)-1; i++) {
            if (S[i]==S[i+1]){
                duptmp++;
            }
            else {
                idup += duptmp/2;
                duptmp = 1;
            }
        }
        if (duptmp>0) {
            edup = duptmp;
        }
    }
//    printf("t %lld i %lld e %lld tmp %lld\n", tdup, idup, edup, duptmp);

    if (tdup==strlen(S)) {
        printf("%lld\n", tdup*K/2);
    }
    else if (S[0]!=S[strlen(S)-1])
        printf("%lld\n", (tdup/2+idup+edup/2)*K);
    else {
//        printf("%lld %lld %lld %lld\n", tdup/2 ,idup * K, (tdup+edup)/2*(K-1) ,edup/2);
        printf("%lld\n", tdup/2 + idup * K + (tdup+edup)/2*(K-1) + edup/2);
    }

    return 0;
}