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.StringTokenizer;

/**
 * 173A
 * θ(min(n, lcm(m,k))) time
 * θ(m + k) space
 *
 * @author artyom
 */
public class _173A implements Runnable {
    private BufferedReader in;
    private StringTokenizer tok;

    private Object solve() throws IOException {
        int n = nextInt();
        String a = nextToken(), b = nextToken();
        int m = a.length(), k = b.length(), lcm = (k * m) / gcd(k, m), r = n % lcm, ac = 0, bc = 0, ar = 0, br = 0;
        for (int i = 0, lim = Math.min(n, lcm); i < lim; i++) {
            char ax = a.charAt(i % m), bx = b.charAt(i % k);
            if (ax == 'R') {
                if (bx == 'S') {
                    ac++;
                } else if (bx == 'P') {
                    bc++;
                }
            } else if (ax == 'S') {
                if (bx == 'P') {
                    ac++;
                } else if (bx == 'R') {
                    bc++;
                }
            } else {
                if (bx == 'R') {
                    ac++;
                } else if (bx == 'S') {
                    bc++;
                }
            }
            if (i + 1 == r) {
                ar = ac;
                br = bc;
            }
        }
        ac = ac * (n / lcm) + ar;
        bc = bc * (n / lcm) + br;
        return bc + " " + ac;
    }

    private static int gcd(int x, int y) {
        return y == 0 ? x : gcd(y, x % y);
    }

    //--------------------------------------------------------------
    public static void main(String[] args) {
        new _173A().run();
    }

    @Override
    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            tok = null;
            System.out.print(solve());
            in.close();
        } catch (IOException e) {
            System.exit(0);
        }
    }

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

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