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.PrintWriter;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(System.out, true);
        Scanner in = new Scanner(System.in);
        Main solver = new Main(in, out);
        solver.run();
        in.close();
        out.close();
        return;
    }

    Scanner in;
    PrintWriter out;

    public Main(Scanner i, PrintWriter o) {
        in = i;
        out = o;
    }

    int gcd(int a, int b) {
        if (a < b) {
            int t = a;
            a = b;
            b = t;
        }
        int c;
        while (b > 0) {
            c = a % b;
            a = b;
            b = c;
        }
        return a;
    }

    int common(int m, int n) {
        int a = gcd(m, n);
        return m * n / a;
    }

    public void run() {
        int R = in.nextInt();
        char[] s1 = in.next().toCharArray();
        char[] s2 = in.next().toCharArray();
        int m = s1.length;
        int n = s2.length;

        int N = common(m, n);
        int i = 0, j = 0, r = 0;
        int win1 = 0, win2 = 0;
        while (r < N) {
            int t = comp(s1[i], s2[j]);
            if (t == 1) {
                win1++;
            }
            else if (t == -1) {
                win2++;
            }
            i = (i+1) % m;
            j = (j+1) % n;
            r++;
        }

        int u = R / N;
        int left = R % N;
        int sum1 = u*win1;
        int sum2 = u*win2;

        r = 0;
        i = j = 0;
        while (r < left) {
            int t = comp(s1[i], s2[j]);
            if (t == 1) {
                sum1++;
            }
            else if (t == -1) {
                sum2++;
            }
            i = (i+1) % m;
            j = (j+1) % n;
            r++;
        }
        out.format("%d %d\n", sum2, sum1);
        return;
    }

    int comp(char a, char b) {
        if (a == b) return 0;
        if (a == 'R' && b == 'S') return 1;
        if (a == 'S' && b == 'P') return 1;
        if (a == 'P' && b == 'R') return 1;
        return -1;
    }

}