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.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;

public class A173 {

    static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static PrintWriter out = new PrintWriter(System.out);
    
    static int nextInt() throws Exception {
        in.nextToken();
        return (int)in.nval;
    }
    
    static String nextString() throws Exception {
        in.nextToken();
        return in.sval;
    }
    
    public static void main(String[] args) throws Exception {
        int n = nextInt();
        char[] a = nextString().toCharArray();
        char[] b = nextString().toCharArray();
        int na = a.length, nb = b.length;
        int m = na*nb/gcd(na, nb);
        int[] sum1 = new int[m];
        int[] sum2 = new int[m];
        sum1[0] = wins(a[0], b[0]);
        sum2[0] = wins(b[0], a[0]);
        for (int i = 1; i < m; i++) {
            sum1[i] = sum1[i-1] + wins(a[i%na], b[i%nb]);
            sum2[i] = sum2[i-1] + wins(b[i%nb], a[i%na]);
        }
        int tot1 = n/m*sum1[m-1] + (n%m != 0 ? sum1[n%m-1] : 0);
        int tot2 = n/m*sum2[m-1] + (n%m != 0 ? sum2[n%m-1] : 0);
        out.println(tot2 + " " + tot1);
        out.flush();
    }
    
    static int wins(char a, char b) {
        if (a == 'R' && b == 'S') return 1;
        else if (a == 'S' && b == 'P') return 1;
        else if (a == 'P' && b == 'R') return 1;
        else return 0;
    }

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