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;
int inline ABS(int a) {
  if (a > 0)
    return a;
  else
    return -a;
}
int GCD(int a, int b) {
  if (!a) return b;
  return GCD(b % a, a);
}
int LCM(int a, int b) {
  int gcd = GCD(a, b);
  return (a / gcd) * b;
}
int main() {
  int n;
  string ni, po;
  cin >> n;
  cin >> ni >> po;
  int lni = ni.size();
  int lpo = po.size();
  int lcm = LCM(lni, lpo);
  int ini = 0, ipo = 0;
  int ani = 0, apo = 0;
  int c = 0;
  c = n / lcm;
  n = n - (n / lcm) * lcm;
  for (int i = 1; i <= n; i++) {
    if (ni[ini] == 'R') {
      if (po[ipo] == 'R')
        ;
      if (po[ipo] == 'S') apo++;
      if (po[ipo] == 'P') ani++;
    }
    if (ni[ini] == 'S') {
      if (po[ipo] == 'R') ani++;
      if (po[ipo] == 'S')
        ;
      if (po[ipo] == 'P') apo++;
    }
    if (ni[ini] == 'P') {
      if (po[ipo] == 'R') apo++;
      if (po[ipo] == 'S') ani++;
      if (po[ipo] == 'P')
        ;
    }
    ini++;
    if (ini == lni) ini = 0;
    ipo++;
    if (ipo == lpo) ipo = 0;
  }
  int lapo = 0, lani = 0;
  ipo = ini = 0;
  for (int i = 1; i <= lcm; i++) {
    if (ni[ini] == 'R') {
      if (po[ipo] == 'R')
        ;
      if (po[ipo] == 'S') lapo++;
      if (po[ipo] == 'P') lani++;
    }
    if (ni[ini] == 'S') {
      if (po[ipo] == 'R') lani++;
      if (po[ipo] == 'S')
        ;
      if (po[ipo] == 'P') lapo++;
    }
    if (ni[ini] == 'P') {
      if (po[ipo] == 'R') lapo++;
      if (po[ipo] == 'S') lani++;
      if (po[ipo] == 'P')
        ;
    }
    ini++;
    if (ini == lni) ini = 0;
    ipo++;
    if (ipo == lpo) ipo = 0;
  }
  apo = apo + c * lapo;
  ani = ani + c * lani;
  cout << ani << " " << apo << "\n";
  return 0;
}
