// Khajiit // Solution by Jacob Plachta #define DEBUG 0 #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define LL long long #define LD long double #define PR pair #define Fox(i,n) for (i=0; i=0; i--) #define FoxR1(i,n) for (i=n; i>0; i--) #define FoxRI(i,a,b) for (i=b; i>=a; i--) #define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++) #define Min(a,b) a=min(a,b) #define Max(a,b) a=max(a,b) #define Sz(s) int((s).size()) #define All(s) (s).begin(),(s).end() #define Fill(s,v) memset(s,v,sizeof(s)) #define pb push_back #define mp make_pair #define x first #define y second template T Abs(T x) { return(x < 0 ? -x : x); } template T Sqr(T x) { return(x * x); } string plural(string s) { return(Sz(s) && s[Sz(s) - 1] == 'x' ? s + "en" : s + "s"); } const int INF = (int)1e9; const LD EPS = 1e-12; const LD PI = acos(-1.0); #if DEBUG #define GETCHAR getchar #else #define GETCHAR getchar_unlocked #endif bool Read(int& x) { char c, r = 0, n = 0; x = 0; for (;;) { c = GETCHAR(); if ((c < 0) && (!r)) return(0); if ((c == '-') && (!r)) n = 1; else if ((c >= '0') && (c <= '9')) x = x * 10 + c - '0', r = 1; else if (r) break; } if (n) x = -x; return(1); } bool ReadLL(LL& x) { char c, r = 0, n = 0; x = 0; for (;;) { c = GETCHAR(); if ((c < 0) && (!r)) return(0); if ((c == '-') && (!r)) n = 1; else if ((c >= '0') && (c <= '9')) x = x * 10 + c - '0', r = 1; else if (r) break; } if (n) x = -x; return(1); } #define MOD 1000000007 #define LIM 1000005 int N, M; char X[LIM], Y[LIM]; LL Solve(int s) { int i; LL sum = 0; int hasA = 0, needsA = 0; FoxR(i, M) { if (X[s + i] == 'A') hasA++; if (Y[s + i] == 'A') needsA++; sum += Abs(hasA - needsA); } return(sum); } int main() { if (DEBUG) freopen("in.txt", "r", stdin); // vars int T, t; int i; // testcase loop Read(T); Fox1(t, T) { // input Read(N), Read(M); scanf("%s%s", &X, &Y); // handle each spoke LL ans = 0; Fox(i, N) ans += Solve(i * M + 1); // output printf("Case #%d: %lld\n", t, ans); } return(0); }