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.
import java.io.*;
import java.util.*;

public class Main {
    public static final Reader reader = new InputStreamReader(System.in);
    public static final Writer writer = new OutputStreamWriter(System.out);
    public static final BufferedReader br = new BufferedReader(reader);
    public static final StreamTokenizer st = new StreamTokenizer(br);
    public static final PrintWriter out = new PrintWriter(writer);
    public static final Scanner in = new Scanner(reader);
    public static int nextInt() throws IOException {
        st.nextToken();
        return (int)(st.nval);
    }
        
    public static void main(String[] args) throws IOException {
        int n = in.nextInt();
        String str1 = in.next();
        int len1 = str1.length();
        String str2 = in.next();        
        int len2 = str2.length();
        int win1 = 0, win2 = 0;
        int k = len1 * len2;
        for (int i = 0; i < k; ++i) {
            int i1 = i % len1;
            int i2 = i % len2;
            if (str1.charAt(i1) == 'S' && str2.charAt(i2) == 'P') win1 += (n / k) + (i < (n % k) ? 1 : 0);
            if (str1.charAt(i1) == 'P' && str2.charAt(i2) == 'R') win1 += (n / k) + (i < (n % k) ? 1 : 0);
            if (str1.charAt(i1) == 'R' && str2.charAt(i2) == 'S') win1 += (n / k) + (i < (n % k) ? 1 : 0);

            if (str2.charAt(i2) == 'S' && str1.charAt(i1) == 'P') win2 += (n / k) + (i < (n % k) ? 1 : 0);
            if (str2.charAt(i2) == 'P' && str1.charAt(i1) == 'R') win2 += (n / k) + (i < (n % k) ? 1 : 0);
            if (str2.charAt(i2) == 'R' && str1.charAt(i1) == 'S') win2 += (n / k) + (i < (n % k) ? 1 : 0);
        }
        out.println(win2 + " " + win1);
        out.flush();
    }
}
