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)

#include <bits/stdc++.h>
#define GET_MACRO(_1,_2,_3,_4,_5,_6,_7,_8,NAME,...) NAME
#define pr(...) cerr<< GET_MACRO(__VA_ARGS__,pr8,pr7,pr6,pr5,pr4,pr3,pr2,pr1)(__VA_ARGS__) <<endl
#define pr1(a) (#a)<<"="<<(a)<<" "
#define pr2(a,b) pr1(a)<<pr1(b)
#define pr3(a,b,c) pr1(a)<<pr2(b,c)
#define pr4(a,b,c,d) pr1(a)<<pr3(b,c,d)
#define pr5(a,b,c,d,e) pr1(a)<<pr4(b,c,d,e)
#define pr6(a,b,c,d,e,f) pr1(a)<<pr5(b,c,d,e,f)
#define pr7(a,b,c,d,e,f,g) pr1(a)<<pr6(b,c,d,e,f,g)
#define pr8(a,b,c,d,e,f,g,h) pr1(a)<<pr7(b,c,d,e,f,g,h)
#define prArr(a) {cerr<<(#a)<<"={";int i=0;for(auto t:(a))cerr<<(i++?", ":"")<<t;cerr<<"}"<<endl;}
using namespace std;
using Int = long long;
using _int = int;
using ll = long long;
using Double = long double;
const Int INF = (1LL<<60)+1e9; // ~ 1.15 * 1e18
const Int mod = (1e9)+7;
const Double EPS = 1e-8;
const Double PI = 6.0 * asin((Double)0.5);
using P = pair<Int,Int>;
template<class T> T Max(T &a,T b){return a=max(a,b);}
template<class T> T Min(T &a,T b){return a=min(a,b);}
template<class T1, class T2> ostream& operator<<(ostream& o,pair<T1,T2> p){return o<<"("<<p.first<<","<<p.second<<")";}
template<class T1, class T2, class T3> ostream& operator<<(ostream& o,tuple<T1,T2,T3> t){
  return o<<"("<<get<0>(t)<<","<<get<1>(t)<<","<<get<2>(t)<<")";}
template<class T1, class T2> istream& operator>>(istream& i,pair<T1,T2> &p){return i>>p.first>>p.second;}
template<class T> ostream& operator<<(ostream& o,vector<T> a){Int i=0;for(T t:a)o<<(i++?" ":"")<<t;return o;}
template<class T> istream& operator>>(istream& i,vector<T> &a){for(T &t:a)i>>t;return i;}
//INSERT ABOVE HERE

Int calc(string s, Int K){

  Int N = s.size();
  vector<vector<Int> > dp(K * N + 1, vector<Int>(3, INF)); //[回数 * 添字][前回の文字(0:同じ, 1:変えた)]

  dp[0][1] = 0;
  for(Int i=0;i<K * N;i++){
    {//k == 0
      if(i == 0 || s[i%N] != s[(i-1 + N)%N]) Min(dp[i+1][0], dp[i][0]); //変えない
      Min(dp[i+1][1], dp[i][0] + 1); //変える
    }
    {//k == 1
      Min(dp[i+1][0], dp[i][1]);
      Min(dp[i+1][1], dp[i][1]+1);
    }
  }
  Int res = min(dp[K * N][0], dp[K * N][1]);
  return res;
}

signed main(){
  srand((unsigned)time(NULL));
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(12);

  string s;
  cin>>s;

  Int K;
  cin>>K;

  if(K < 10){
    Int ans = calc(s, K);
    cout<<ans<<endl;
    return 0;
  }

  Int a = calc(s, 4);
  Int b = calc(s, 5);
  Int ans = a + (b - a) * (K - 4);
  cout<<ans<<endl;


  return 0;
}