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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0, i##_len = (n); i < i##_len; i++)
#define reps(i, s, n) for(int i = (s), i##_len = (n); i < i##_len; i++)
#define rrep(i, n) for(int i = (n) - 1; i >= 0; i--)
#define rreps(i, e, n) for(int i = (n) - 1; i >= (e); i--)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((int)(x).size())
#define uniq(x) (x).erase(unique((x).begin(), (x).end()), (x).end())

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    string s;
    int k;
    cin >> s >> k;
    int sl = s.length();
    int fc = 0, lc = 0, cc = 0, nc = 0;
    bool bf = false;
    ll ans = 0;
    
    rep(i, sl) {
        if (s[0] == s[i]) {
            fc++;
        }
        else {
            break;
        }
    }
    
    rrep(i, sl) {
        if (s[0] == s[i]) {
            lc++;
        }
        else {
            break;
        }
    }
    
    if (fc == sl) {
        ans = (ll)sl * k / 2;
    }
    else {
        rep(i, sl - 1) {
            if ((!bf) && (s[i] == s[i + 1])) {
                cc++;
                bf = true;
            }
            else {
                bf = false;
            }
        }
        
        if ((fc % 2 == 1) && (lc % 2 == 1)) {
            nc++;
        }
        
        ans = (ll)cc * k + (ll)nc * (k - 1);
    }
    
    cout << ans << endl;
    return 0;
}