Nikephoros and Polycarpus play rock-paper-scissors. The loser gets pinched (not too severely!).

Let us remind you the rules of this game. Rock-paper-scissors is played by two players. In each round the players choose one of three items independently from each other. They show the items with their hands: a rock, scissors or paper. The winner is determined by the following rules: the rock beats the scissors, the scissors beat the paper and the paper beats the rock. If the players choose the same item, the round finishes with a draw.

Nikephoros and Polycarpus have played n rounds. In each round the winner gave the loser a friendly pinch and the loser ended up with a fresh and new red spot on his body. If the round finished in a draw, the players did nothing and just played on.

Nikephoros turned out to have worked out the following strategy: before the game began, he chose some sequence of items A = (a1, a2, ..., am), and then he cyclically showed the items from this sequence, starting from the first one. Cyclically means that Nikephoros shows signs in the following order: a1, a2, ..., am, a1, a2, ..., am, a1, ... and so on. Polycarpus had a similar strategy, only he had his own sequence of items B = (b1, b2, ..., bk).

Determine the number of red spots on both players after they've played n rounds of the game. You can consider that when the game began, the boys had no red spots on them.

Input

The first line contains integer n (1 ≤ n ≤ 2·109) — the number of the game's rounds.

The second line contains sequence A as a string of m characters and the third line contains sequence B as a string of k characters (1 ≤ m, k ≤ 1000). The given lines only contain characters "R", "S" and "P". Character "R" stands for the rock, character "S" represents the scissors and "P" represents the paper.

Output

Print two space-separated integers: the numbers of red spots Nikephoros and Polycarpus have.

Examples

Input

7
RPS
RSPP


Output

3 2

Input

5
RRRRRRRR
R


Output

0 0

Note

In the first sample the game went like this:

  * R - R. Draw. 
  * P - S. Nikephoros loses. 
  * S - P. Polycarpus loses. 
  * R - P. Nikephoros loses. 
  * P - R. Polycarpus loses. 
  * S - S. Draw. 
  * R - P. Nikephoros loses. 



Thus, in total Nikephoros has 3 losses (and 3 red spots), and Polycarpus only has 2.
#include <bits/stdc++.h>
using namespace std;
template <class F, class T>
T convert(F a, int p = -1) {
  stringstream ss;
  if (p >= 0) ss << fixed << setprecision(p);
  ss << a;
  T r;
  ss >> r;
  return r;
}
template <class T>
void db(T a, int p = -1) {
  if (p >= 0) cout << fixed << setprecision(p);
  cout << a << " ";
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T cube(T x) {
  return x * x * x;
}
template <class T>
struct Triple {
  T x, y, z;
  Triple() {}
  Triple(T _x, T _y, T _z) : x(_x), y(_y), z(_z) {}
};
template <class T>
Triple<T> euclid(T a, T b) {
  if (b == 0) return Triple<T>(1, 0, a);
  Triple<T> r = euclid(b, a % b);
  return Triple<T>(r.y, r.x - a / b * r.y, r.z);
}
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return s == 0 ? 0 : cntbit(s >> 1) + (s & 1);
}
const int bfsz = 1 << 16;
char bf[bfsz + 5];
int rsz = 0;
int ptr = 0;
char gc() {
  if (rsz <= 0) {
    ptr = 0;
    rsz = fread(bf, 1, bfsz, stdin);
    if (rsz <= 0) return EOF;
  }
  --rsz;
  return bf[ptr++];
}
void ga(char &c) {
  c = EOF;
  while (!isalpha(c)) c = gc();
}
int gs(char s[]) {
  int l = 0;
  char c = gc();
  while (isspace(c)) c = gc();
  while (c != EOF && !isspace(c)) {
    s[l++] = c;
    c = gc();
  }
  s[l] = '\0';
  return l;
}
template <class T>
bool gi(T &v) {
  v = 0;
  char c = gc();
  while (c != EOF && c != '-' && !isdigit(c)) c = gc();
  if (c == EOF) return false;
  bool neg = c == '-';
  if (neg) c = gc();
  while (isdigit(c)) {
    v = v * 10 + c - '0';
    c = gc();
  }
  if (neg) v = -v;
  return true;
}
const double PI = 2 * acos(0);
const string months[] = {"January",   "February", "March",    "April",
                         "May",       "June",     "July",     "August",
                         "September", "October",  "November", "December"};
const int days[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int dr[] = {-1, 0, +1, 0};
const int dc[] = {0, +1, 0, -1};
const long long oo = 100000000000000005LL;
const double eps = 1e-9;
int n, m, k;
string a, b;
int which(char c1, char c2) {
  if (c1 == c2) return 0;
  if (c1 == 'R') return c2 == 'S' ? +1 : -1;
  if (c1 == 'S') return c2 == 'P' ? +1 : -1;
  return c2 == 'R' ? +1 : -1;
}
int main() {
  cin >> n;
  cin >> a >> b;
  m = a.length();
  k = b.length();
  int l = lcm(m, k);
  int ca = 0, cb = 0;
  for (typeof(l) i = 0; i < (l); ++i) {
    int r = which(a[i % m], b[i % k]);
    if (r == 0) continue;
    if (r == +1)
      ++cb;
    else
      ++ca;
  }
  int T = n / l;
  int D = n % l;
  int xa = 0, xb = 0;
  for (typeof(D) i = 0; i < (D); ++i) {
    int r = which(a[i % m], b[i % k]);
    if (r == 0) continue;
    if (r == +1)
      ++xb;
    else
      ++xa;
  }
  int ra = ca * T + xa;
  int rb = cb * T + xb;
  cout << ra << " " << rb << endl;
  return 0;
}
