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.util.Scanner;


public class A {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String s = sc.next();
        String s2 = sc.next();
        int k =(s.length()*s2.length())/gcd(s.length(),s2.length());
        String x = s;
        String x1 = s2;
        int t = x.length();
        for (int i = 1; i <k/t; i++) {
            s =s.concat(x);
        }
        for (int i = 1; i <k/x1.length(); i++) {
            s2 =s2.concat(x1);
        }
        int ni = 0;
        int pi = 0;
        if (n >= k){
            for (int i = 0; i <k; i++) {
            if ((s.charAt(i) =='P' && s2.charAt(i) =='S') || (s.charAt(i) =='R' && s2.charAt(i) =='P')||(s.charAt(i) =='S' && s2.charAt(i) =='R')){
                ni++;
            }else
            if ((s.charAt(i) =='S' && s2.charAt(i) =='P') || (s.charAt(i) =='P' && s2.charAt(i) =='R')||(s.charAt(i) =='R' && s2.charAt(i) =='S') ){
                        pi++;
                    }
           }
            ni *=n/k;
            pi *=n/k;
            for (int i = 0; i < n%k; i++) {
                if ((s.charAt(i) =='P' && s2.charAt(i) =='S') ||
                 (s.charAt(i) =='R' && s2.charAt(i) =='P')||
                 (s.charAt(i) =='S' && s2.charAt(i) =='R')){
                    ni++;
                }else
                if ((s.charAt(i) =='S' && s2.charAt(i) =='P') ||
                         (s.charAt(i) =='P' && s2.charAt(i) =='R')||
                         (s.charAt(i) =='R' && s2.charAt(i) =='S') ){
                            pi++;
                        }
            }
        }else{
            for (int i = 0; i <n; i++) {
            if ((s.charAt(i) =='P' && s2.charAt(i) =='S') ||
             (s.charAt(i) =='R' && s2.charAt(i) =='P')||
             (s.charAt(i) =='S' && s2.charAt(i) =='R')){
                ni++;
            }
            if ((s.charAt(i) =='S' && s2.charAt(i) =='P') ||
                     (s.charAt(i) =='P' && s2.charAt(i) =='R')||
                     (s.charAt(i) =='R' && s2.charAt(i) =='S') ){
                        pi++;
                    }
        }   
        }
        System.out.println(ni+" "+pi);
    }

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

}
