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 int gcd(int a, int b) {
        if (a == 0 || b == 0) {
            return a + b;
        } else if (a > b) {
            return gcd(b, a % b);
        } else {
            return gcd(a, b % a);
        }
    }

    public static void main(String args[]) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        in.nextLine();
        String f = in.nextLine();
        String s = in.nextLine();
        int p1 = 0;
        int p2 = 0;
        int t1 = 0;
        int t2 = 0;

        int per = (f.length() * s.length()) / gcd(f.length(), s.length());
        if (per <= n) {
            int a1 = 0;
            int a2 = 0;
            for (int i = 0; i < per; i++) {
                if (f.charAt(p1) == 'S') {
                    if (s.charAt(p2) == 'P') {
                        a2++;
                    } else if (s.charAt(p2) == 'R') {
                        a1++;
                    }
                } else if (f.charAt(p1) == 'P') {
                    if (s.charAt(p2) == 'S') {
                        a1++;
                    } else if (s.charAt(p2) == 'R') {
                        a2++;
                    }
                } else if (f.charAt(p1) == 'R') {
                    if (s.charAt(p2) == 'S') {
                        a2++;
                    } else if (s.charAt(p2) == 'P') {
                        a1++;
                    }
                }
                p1++;
                p1 %= f.length();
                p2++;
                p2 %= s.length();
            }
            t1= (n/per) * a1;
            t2 = (n/per) * a2;
            n = n % per;
        }
        p1 = p2 = 0;
        for (int i = 0; i < n; i++) {
            if (f.charAt(p1) == 'S') {
                if (s.charAt(p2) == 'P') {
                    t2++;
                } else if (s.charAt(p2) == 'R') {
                    t1++;
                }
            } else if (f.charAt(p1) == 'P') {
                if (s.charAt(p2) == 'S') {
                    t1++;
                } else if (s.charAt(p2) == 'R') {
                    t2++;
                }
            } else if (f.charAt(p1) == 'R') {
                if (s.charAt(p2) == 'S') {
                    t2++;
                } else if (s.charAt(p2) == 'P') {
                    t1++;
                }
            }
            p1++;
            p1 %= f.length();
            p2++;
            p2 %= s.length();
        }
        System.out.println(t1+" "+t2);
    }
}
