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.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;

public class A {
    
    private BufferedReader in;
    private PrintWriter out;
    
    final long max = 2000000005;
    
    public static void main (String[] args) throws IOException {
        new A().solve();
    }
    
    void solve () throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(new OutputStreamWriter(System.out));
        
        int n = new Integer(in.readLine());
        String a = in.readLine();
        String b = in.readLine();
        
        int l = lcm(a.length(), b.length());
        
        int ii = 0, jj = 0;
        ArrayList<Character> ar = new ArrayList<Character>();
        ArrayList<Character> br = new ArrayList<Character>();
        int aa = 0, bb = 0;
        for (int i = 0; i < l; i++) {
            ar.add(a.charAt(ii));
            br.add(b.charAt(jj));
            ii = (ii + 1) % a.length();
            jj = (jj + 1) % b.length();
        }
        
        for (int i = 0; i < Math.min(n, l); i++) {
            int r = get(ar.get(i), br.get(i));
            if (r == 1) {
                aa++;
            } else if (r == -1) {
                bb++;
            }
        }
        if (n <= l) {
            out.print(aa + " " + bb);
            
            out.flush();
            return;
        }
        
        int d = n / l;
        if (n % l == 0) {
            out.print(aa * d + " " + bb * d);
            out.flush();
            return;
        }
        
        aa *= d;
        bb *= d;
        for (int i = 0; i< n % l; i++) {
            int r = get(ar.get(i), br.get(i));
            if (r == 1) {
                aa++;
            } else if (r == -1) {
                bb++;
            }
        }
        
        out.print(aa + " " + bb);
        
        out.flush();
    }
    
    int get(char a, char b) {
        if (a== 'R' && b== 'S') {
            return -1;
        } else if (b== 'R' && a== 'S') {
            return 1;
        } else if (a== 'R' && b== 'P') {
            return 1;
        } else if (b== 'R' && a== 'P') {
            return -1;
        } else if (a== 'S' && b== 'P') {
            return -1;
        } else if (b== 'S' && a== 'P') {
            return 1;
        }
        return 0;
    }
    
    int nod_recur(int num1, int num2)
    {
      int n = num1 % num2;
      num1 = num2;
      num2 = n;
      if (n > 0)
        return nod_recur(num1, num2);
      else
        return num1;
    }
    
    int lcm (int a, int b) {
        return a / nod_recur (a, b) * b;
    }

}
