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;
import java.io.PrintWriter;
import java.util.Hashtable;


public class Main {

    /**
     * @param args
     * @throws IOException 
     */
    public static void main(String[] args) throws IOException {
        // TODO Auto-generated method stub
        BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
        //Scanner in=new Scanner(System.in);
        PrintWriter out=new PrintWriter(System.out);
        Integer n=Integer.parseInt(in.readLine());
        Integer nRem=n;
        char[] f=in.readLine().toCharArray();
        char[] s=in.readLine().toCharArray();
        int fLen=f.length;
        int sLen=s.length;
        int i=0;
        int j=0;
        int count1=0;
        int count2=0;
        while(n>0){
            if(f[i]=='R' && s[j]=='S') count1++;
            else if(f[i]=='S' && s[j]=='P') count1++;
            else if(f[i]=='P' && s[j]=='R') count1++;
            else if(s[j]=='R' && f[i]=='S') count2++;
            else if(s[j]=='S' && f[i]=='P') count2++;
            else if(s[j]=='P' && f[i]=='R') count2++;
            n--;
            i++;
            if(i==fLen) i=0;
            j++;
            if(j==sLen) j=0;
            if(n!=0 && i==0 && j==0){
                int mod=nRem/(nRem-n);
                count1*=mod;
                count2*=mod;
                n=nRem%(nRem-n);
            }
        }
        out.print(count2+" "+count1);
        out.close();
    }

}
