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 static java.lang.Double.parseDouble;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.System.exit;

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

public class RockPaperScissors {

    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;

    static int nextInt() throws IOException {
        return parseInt(next());
    }

    static long nextLong() throws IOException {
        return parseLong(next());
    }

    static double nextDouble() throws IOException {
        return parseDouble(next());
    }

    static String next() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    public static void main(String[] args) {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(new OutputStreamWriter(System.out));
            solve();
            in.close();
            out.close();
        } catch (Throwable e) {
            e.printStackTrace();
            exit(1);
        }
    }

    private static void solve() throws IOException {
        int n = nextInt();
        String a = next();
        String b = next();
        int an = a.trim().length();
        int bn = b.trim().length();
        int g = gcd(an, bn);
        int s = an*bn/g;
        int ar = 0;
        int br = 0;
        int res1 = 0, res2 = 0;
        for (int i = 0; i < s; i ++) {
            char ac = a.charAt(i%an);
            char bc = b.charAt(i%bn);
            if (ac == bc) continue;
            if (ac == 'R' && bc == 'S' || ac == 'S' && bc == 'P' || ac == 'P' && bc == 'R') {
                ar ++;
                if (i < (n%s)) res1 ++;
            } else {
                br ++;
                if (i < (n%s)) res2 ++;
            }
        }
        res1 += (n/s)*ar;
        res2 += (n/s)*br;
        System.out.println(res2 + " " + res1);
    }

    private static int gcd(int a, int b) {
        if (b != 0)
            return gcd(b, a % b);
        else
            return a;
    }

}