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;
bool DBG;
struct dbg {
  template <typename T>
  dbg& operator,(const T& v) {
    if (DBG) cerr << v << " ";
    return *this;
  }
  ~dbg() {
    if (DBG) cerr << endl;
  }
};
unsigned int gcd(unsigned int a, unsigned int b) {
  unsigned int c;
  while (b) {
    c = a % b;
    a = b;
    b = c;
  }
  return a;
}
unsigned int lcm(unsigned int a, unsigned int b) {
  return ((a / gcd(a, b)) * b);
}
map<char, map<char, int> > RPS;
int main(int argc, char* argv[]) {
  if (argc > 1) DBG = ((*argv[1] == '0') ? 0 : 1);
  RPS['P']['R'] = 1;
  RPS['R']['P'] = -1;
  RPS['R']['S'] = 1;
  RPS['S']['R'] = -1;
  RPS['S']['P'] = 1;
  RPS['P']['S'] = -1;
  RPS['P']['P'] = 0;
  RPS['R']['R'] = 0;
  RPS['S']['S'] = 0;
  unsigned int n;
  scanf("%u", &n);
  string N, P;
  cin >> N >> P;
  unsigned int m = N.length(), k = P.length(), p = lcm(m, k), spotsN = 0,
               spotsP = 0, c = ((n < p) ? n : p);
  unsigned int sp = 0, sn = 0;
  for (int i = 0; i < c; i++) {
    if (RPS[N[i % m]][P[i % k]] == 1)
      sp++;
    else if (RPS[N[i % m]][P[i % k]] == -1)
      sn++;
  }
  spotsP = (n / p) * sp;
  spotsN = (n / p) * sn;
  for (int i = 0; i < n % p; i++) {
    if (RPS[N[i % m]][P[i % k]] == 1)
      spotsP++;
    else if (RPS[N[i % m]][P[i % k]] == -1)
      spotsN++;
  }
  printf("%u %u\n", spotsN, spotsP);
  return 0;
}
