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.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Solution {

    static final int beats(char x, char y) {
        if (x == y) return 0;
        if (x == 'R' && y == 'S') return 1;
        if (x == 'S' && y == 'P') return 1;
        if (x == 'P' && y == 'R') return 1;
        return -1;
    }
    
    public static void main(String[] args) {
        FastScanner in = new FastScanner(System.in);
        int n = in.nextInt();
        String a = in.next(), b = in.next();
        int m = a.length(), k = b.length();
        int nic = 0, pol = 0;
        int[] ha = new int[m * k + 1], hb = new int[m * k + 1];
        int pa = 0, pb = 0;
        for (int i = 1; i <= m * k; i++) {
            int outcome = beats(a.charAt(pa), b.charAt(pb));
            if (outcome < 0) nic++;
            else if (outcome > 0) pol++;
            ha[i] = nic;
            hb[i] = pol;
            pa = (pa + 1) % m;
            pb = (pb + 1) % k;
        }
        int q = n / (m * k), r = n % (m * k);
        System.out.println((nic * q + ha[r]) + " " + (pol * q + hb[r]));
    }

}

final class FastScanner {
    private StringTokenizer stringTokenizer = null;
    private BufferedReader bufferedReader; 
    
    public FastScanner(InputStream in) {
        bufferedReader = new BufferedReader(new InputStreamReader(in));
        try {
            stringTokenizer = new StringTokenizer(bufferedReader.readLine());
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    
    public String next() {
        while (!stringTokenizer.hasMoreTokens())
            try {
                stringTokenizer = new StringTokenizer(bufferedReader.readLine());
            } catch (IOException e) {
                e.printStackTrace();
                return null;
            }
        return stringTokenizer.nextToken();
    }
    
    public int nextInt() {
        return Integer.parseInt(next());
    }
    
    public long nextLong() {
        return Long.parseLong(next());
    }
    
    public double nextDouble() {
        return Double.parseDouble(next());
    }

    public boolean hasNext() {
        return stringTokenizer.hasMoreTokens();
    }
    
    public void nextLine() {
        try {
            stringTokenizer = new StringTokenizer(bufferedReader.readLine());
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}
