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| #include <bits/stdc++.h>
#pragma optimization_level 3 #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3") #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx") #pragma GCC optimize("Ofast") #pragma GCC target("avx,avx2,fma") #pragma GCC optimization ("unroll-loops")
using namespace std;
struct PairHash {inline std::size_t operator()(const std::pair<int, int> &v) const { return v.first * 31 + v.second; }};
#define Code ios_base::sync_with_stdio(false); #define By ios::sync_with_stdio(0); #define Sumfi cout.tie(NULL);
using ll = long long; using ld = long double;
const ld PI = 3.14159265358979323846; const ll INF = 1e18; const ld EPS = 1e-9; const ll MAX_N = 101010; const ll mod = 998244353;
typedef pair<ll, ll> pll; typedef vector<pll> vpll; typedef array<ll,3> all3; typedef array<ll,5> all5; typedef vector<all3> vall3; typedef vector<all5> vall5; typedef vector<ld> vld; typedef vector<ll> vll; typedef vector<vll> vvll; typedef vector<int> vi; typedef vector<string> vs; typedef vector<vs> vvs; typedef unordered_set<ll> usll; typedef unordered_set<pll, PairHash> uspll; typedef unordered_map<ll, ll> umll; typedef unordered_map<pll, ll, PairHash> umpll;
#define rep(i,m,n) for(ll i=m;i<n;i++) #define rrep(i,m,n) for(ll i=n;i>=m;i--) #define all(a) begin(a), end(a) #define rall(a) rbegin(a), rend(a) #define ZERO(a) memset(a,0,sizeof(a)) #define MINUS(a) memset(a,0xff,sizeof(a)) #define sz(x) ll((x).size()) #define pyes cout<<"YES\n"; #define pno cout<<"NO\n"; #define pneg1 cout<<"-1\n"; #define CASE(x) cout<<"Case #"<<x<<": ";
template <typename T> void print(T &&t) { cout << t << "\n"; } template<typename T> void printv(vector<T>v){ll n=v.size();rep(i,0,n)cout<<v[i]<<" ";cout<<"\n";}
ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); } ll __lcm(ll x, ll y) { return x * y / __gcd(x,y); } ll modpow(ll n, ll x, ll MOD = mod) { if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }
ll solve(string s, ll k) { vll A; ll l = 0, r = 0, n = sz(s); while(l < n) { while(r < n and s[l] == s[r]) r++; A.push_back(r-l); l = r; } ll res = 0; if(s[0] != s[n-1]) { rep(i,0,sz(A)) res += A[i]/2; return res * k; } else { if(sz(A) == 1) return A[0] * k / 2; l = A[0], r = A.back(); rep(i,1,sz(A)-1) res += A[i]/2; return res * k + (l + r) / 2 * (k - 1) + l / 2 + r / 2; } }
int main() { Code By Sumfi
cout.precision(12);
ll tc = 1; rep(i,1,tc+1) { string s; cin>>s; ll k; cin>>k; print(solve(s,k)); } return 0; }
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