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: Wrong Answer)

#pragma gcc optimize("Ofast")
#include <bits/stdc++.h>

using namespace std;
using ll = long long;

#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(V) (V).begin(),(V).end()
#define SORT(V) sort(ALL(V)) //小さい方からソート
#define REV(V) reverse(ALL(V)) //リバース
#define RSORT(V) SORT(V);REV(V) //大きい方からソート
#define NPN(V) next_permutation(ALL(V)) //順列
#define pb(n) push_back(n)
#define endl '\n'
#define Endl '\n'
#define DUMP(x)  cout << #x << " = " << (x) << endl
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define yes(n) cout << ((n) ? "yes" : "no"  ) << endl
#define Yay(n) cout << ((n) ? "Yay!": ":(") << endl
#define RAPID cin.tie(0);ios::sync_with_stdio(false)
#define VSUM(V) accumulate(ALL(V), 0)
#define MID(a,b,c) (a) < (b) && (b) < (c)
#define MIDe(a,b,c) (a) <= (b) && (b) <= (c)
#define IN(n) cin >> n
#define IN2(a,b) cin >> a >> b
#define IN3(a,b,c) cin >> a >> b >> c 
#define IN4(a,b,c,d) cin >> a >> b >> c >> d
#define VIN(V) for(int i = 0; i < (V).size(); i++) {cin >> (V).at(i);}
#define OUT(n) cout << n << endl
#define VOUT(V) REP(i, (V).size()) {cout << (V)[i] << endl;}
#define VOUT2(V) REP(i, (V).size()) {cout << (V)[i] << " ";} cout<<endl;

// 型マクロ定義
#define int long long
#define P pair<ll,ll>
#define Vi vector<ll>
#define VVi vector<vector<ll>>
#define Vd vector<double>
#define Vb vector<bool>
#define Vs vector<string>
#define Vc vector<char>
#define M map<ll,ll>
#define S set<ll>
#define PQ priority_queue<ll>
#define PQG priority_queue<ll,V,greater<ll>
//

template<class T> inline bool chmin(T& a, T b) {if(a>b){a=b;return true;}return false;}
template<class T> inline bool chmax(T& a, T b) {if(a<b){a=b;return true;}return false;}

const int MOD = 1000000007;
const int INF = 1061109567;
const double EPS = 1e-10;
const double PI  = acos(-1.0);

int gcd(int a,int b){return b != 0 ? gcd(b, a % b) : a;} //最大公約数
int lcm(int a, int b){return a * b / gcd(a, b);}
string toStrUp(string str){char diff = 'A'-'a';REP(i,str.size()) str[i] += diff;return str;}
string gCh(string str, int key){return str.substr(key,1);}

signed main() {
  RAPID;
  // デフォルト変数定義
  int n=0,m=0,a=0,b=0,c=0,d=0,x=0,y=0,z=0;
  string s="",t="";
  //

  // ここから
  IN2(s,x);
  REP(i,s.size()-1){
    if(s[i]==s[i+1]){
      z++;
    }else{
      y += (z+1)/2;
      z = 0;
    }
  }
  if(s[0]==s[s.size()-1]){
    y++;
  }
  OUT(y*x);
}