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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * Created by IntelliJ IDEA.
 * User: Dzianis_Siarheyeu
 * Date: 4/6/12
 * Time: 9:58 AM
 * To change this template use File | Settings | File Templates.
 */

public class A extends Thread {

    public void run() {
        try {
            int n = nextInt();
            char[] s1 = nextToken().toCharArray();
            char[] s2 = nextToken().toCharArray();
            int c1 = 0;
            int c2 = 0;
            int m = s1.length;
            int k = s2.length;
            int common = m*k;            
            for (int i = 0; i < common; i++) {
                char f = s1[i%m];
                char s = s2[i%k];
                if (f=='R'||f=='S'&&s=='R'||s=='S'){
                    if (f=='R'&&s=='S'){
                        c2++;
                    }
                    if(f=='S'&&s=='R'){
                        c1++;
                    }
                }
                if (f=='P'||f=='S'&&s=='P'||s=='S'){
                    if (f=='P'&&s=='S'){
                        c1++;
                    }
                    if(f=='S'&&s=='P'){
                        c2++;
                    }

                }
                if (f=='R'||f=='P'&&s=='R'||s=='P'){
                    if (f=='R'&&s=='P'){
                        c1++;
                    }
                    if(f=='P'&&s=='R'){
                        c2++;
                    }
                }
            }
            int mul = n/common;
            c1*=mul;
            c2*=mul;
            for (int i = mul*common; i<n; i++){
                char f = s1[i%m];
                char s = s2[i%k];
                if (f=='R'||f=='S'&&s=='R'||s=='S'){
                    if (f=='R'&&s=='S'){
                        c2++;
                    }
                    if(f=='S'&&s=='R'){
                        c1++;
                    }
                }
                if (f=='P'||f=='S'&&s=='P'||s=='S'){
                    if (f=='P'&&s=='S'){
                        c1++;
                    }
                    if(f=='S'&&s=='P'){
                        c2++;
                    }

                }
                if (f=='R'||f=='P'&&s=='R'||s=='P'){
                    if (f=='R'&&s=='P'){
                        c1++;
                    }
                    if(f=='P'&&s=='R'){
                        c2++;
                    }
                }
            }
            out.print(c1 + " " + c2);
        } catch (Exception ex) {
            ex.printStackTrace();
            System.exit(999);
        } finally {
            out.flush();
            out.close();
        }
    }


    public static void main(String[] args) {
        new A().run();
    }

    public A() {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        setPriority(Thread.MAX_PRIORITY);
    }

    private BufferedReader in;
    private PrintWriter out;
    private StringTokenizer stringTokenizer;

    private int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    private double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    private long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    private String nextToken() throws IOException {
        if (stringTokenizer == null || !stringTokenizer.hasMoreElements()) {
            stringTokenizer = new StringTokenizer(in.readLine(), "\t\n\r ");
        }
        return stringTokenizer.nextToken();
    }
}