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;

public class RockPaperScissors {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(in.readLine());
        String a = in.readLine();
        String b = in.readLine();
        int N = 0;
        int P = 0;
        int i = 0;
        int j = 0;
        int it = 0;
        if (!a.equals(b)) {
            while (t > 0 && it < a.length() * b.length()) {
                if (a.charAt(i) == 'P' && b.charAt(j) == 'R')
                    P++;
                else if (a.charAt(i) == 'S' && b.charAt(j) == 'P')
                    P++;
                else if (a.charAt(i) == 'R' && b.charAt(j) == 'S')
                    P++;
                else if (a.charAt(i) != b.charAt(j)) {
                    N++;
                }
                i++;
                it++;
                j++;
                i = i % a.length();
                j = j % b.length();
                t--;
            }
            int mod = t % (a.length() * b.length());
            int times = t / (a.length() * b.length());
            P += P * (times );
            N += N * (times );
            it = 0;
            while (it < mod) {
                if (a.charAt(i) == 'P' && b.charAt(j) == 'R')
                    P++;
                else if (a.charAt(i) == 'S' && b.charAt(j) == 'P')
                    P++;
                else if (a.charAt(i) == 'R' && b.charAt(j) == 'S')
                    P++;
                else if (a.charAt(i) != b.charAt(j)) {
                    N++;
                }
                i++;
                j++;
                i = i % a.length();
                j = j % b.length();
                it++;
            }


        }
        System.out.println(N + " " + P);
    }
}


/*
    int t1 = 0, t2 = 0;
    int mod = m * k;
    int last = n % (mod);
    int x = n / mod;
    t1 += x * n1;
    t2 += x * n2;
}

*/