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.*;
import java.io.*;
import java.awt.Point;
import static java.lang.Math.*;

public class P173A {
    static char c(String s, int i) {
        i%=s.length();
        return s.charAt(i);
    }
    public static void main(String[] args) throws Exception {
        Scanner in = new Scanner(System.in);
        long n = in.nextLong();
        String S = in.next();
        String T = in.next();
        int per = S.length()*T.length();
        long[][] P = new long[1+per][2];

        P[0][0] = 0; P[0][1] = 0;
        for(int i=0; i<per; i++) {
            char c1 = c(S,i);
            char c2 = c(T,i);
            if(c1 == c2) {
                P[i+1][0] = P[i][0];
                P[i+1][1] = P[i][1];
            }
            else if((c1=='P' && c2=='S') || (c1=='S' && c2=='R') || (c1=='R' && c2=='P')) {
                P[i+1][0] = P[i][0]+1;
                P[i+1][1] = P[i][1];
            }
            else {
                P[i+1][0] = P[i][0];
                P[i+1][1] = P[i][1]+1;
            }
            //System.out.println(i+" "+c1+" "+c2+" "+P[i+1][0]+" "+P[i+1][1]);
        }

        long a1 = n/per * P[per][0];
        long a2 = n/per * P[per][1];
        n%=per;
        a1 += P[(int)n][0];
        a2 += P[(int)n][1];
        System.out.println(a1+" "+a2);
    }
}
