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;
char ch1[1100000], ch2[1100000];
int get(char ch1, char ch2) {
  if (ch1 == ch2) return 0;
  if (ch1 == 'R' && ch2 == 'S') return 1;
  if (ch1 == 'S' && ch2 == 'P') return 1;
  if (ch1 == 'P' && ch2 == 'R') return 1;
  return -1;
}
int main() {
  int R;
  while (scanf("%d %s %s", &R, ch1, ch2) != EOF) {
    int len1 = strlen(ch1);
    int len2 = strlen(ch2);
    for (int i = (0); i < (len2); i++) {
      for (int j = (0); j < (len1); j++) ch1[i * len1 + j] = ch1[j];
    }
    for (int i = (0); i < (len1); i++) {
      for (int j = (0); j < (len2); j++) ch2[i * len2 + j] = ch2[j];
    }
    int num1 = 0, num2 = 0;
    int g = len1 * len2;
    for (int i = (0); i < (g); i++) {
      int t = get(ch1[i], ch2[i]);
      if (t == 0) continue;
      if (t > 0)
        num2++;
      else
        num1++;
    }
    if (R % g == 0) {
      int m = R / g;
      printf("%d %d\n", num1 * m, num2 * m);
    } else {
      int m = R / g;
      R %= g;
      num1 *= m;
      num2 *= m;
      for (int i = (0); i < (R); i++) {
        int t = get(ch1[i], ch2[i]);
        if (t == 0) continue;
        if (t > 0)
          num2++;
        else
          num1++;
      }
      printf("%d %d\n", num1, num2);
    }
  }
}
