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.math.BigInteger;
public class Krok1A{
    static BufferedReader br;
    public static void main(String args[])throws Exception{
        br=new BufferedReader(new InputStreamReader(System.in));
        long n=toLong();
        char a[]=toStr().toCharArray();
        char b[]=toStr().toCharArray();
        long la=a.length;
        long lb=b.length;
        long lcc=lcm(la,lb);
        int winA[]=new int[(int)lcc];
        int winB[]=new int[(int)lcc];
        for(int i=0;i<lcc;i++){
            int ja=i%(int)la;
            int jb=i%(int)lb;

            int r=awin(a[ja],b[jb]);
            if(i==0){
                if(1==r)
                    winA[i]=1;
                else if(2==r)
                    winB[i]=1;
                else {
                    winA[i]=0;
                    winB[i]=0;
                }
            }
            else{
                if(1==r){
                    winA[i]=winA[i-1]+1;
                    winB[i]=winB[i-1];
                }
                else if(2==r){
                    winB[i]=winB[i-1]+1;
                    winA[i]=winA[i-1];
                }
                else {
                    winA[i]=winA[i-1];
                    winB[i]=winB[i-1];
                }
            }
        }
        long div=n/lcc;
        long rem=n%lcc;
        
        long resB=div*winB[(int)lcc-1];
        long resA=div*winA[(int)lcc-1];

        if(rem>0){
            resA+=winA[(int)rem-1];
            resB+=winB[(int)rem-1];
        }
        System.out.println(resB+" "+resA);


    }
    public static int awin(char a, char b){
        if(a=='R'){
            if(b=='S'){
                return 1;
            }
            else if(b=='P'){
                return 2;
            }
            else return 3;
        }
        else if(a == 'S')
        {
            if(b=='P'){
                return 1;
            }
            else if(b=='R'){
                return 2;
            }
            else return 3;
        }
        else{
            if(b=='R'){
                return 1;
            }
            else if(b=='S'){
                return 2;
            }
            else return 3;
        }
    
    }

    public static long lcm(long a, long b) {
        return (a * b) / gcd(a, b);
    }
    public static long gcd(long a, long b) {
        if (b == 0) {
            return a;
        } else {
            return gcd(b, a % b);
        }
    }


    /****************************************************************/
    public static int[] toIntArray()throws Exception{
       String str[]=br.readLine().split(" ");
       int k[]=new int[str.length];
       for(int i=0;i<str.length;i++){
            k[i]=Integer.parseInt(str[i]);
       }
       return k;
    }
    public static int toInt()throws Exception{
       return Integer.parseInt(br.readLine());
    }
    public static long[] toLongArray()throws Exception{
       String str[]=br.readLine().split(" ");
       long k[]=new long[str.length];
       for(int i=0;i<str.length;i++){
            k[i]=Long.parseLong(str[i]);
       }
       return k;
    }
    public static long toLong()throws Exception{
       return Long.parseLong(br.readLine());
    }
    public static double[] toDoubleArray()throws Exception{
       String str[]=br.readLine().split(" ");
       double k[]=new double[str.length];
       for(int i=0;i<str.length;i++){
            k[i]=Double.parseDouble(str[i]);
       }
       return k;
    }
    public static double toDouble()throws Exception{
       return Double.parseDouble(br.readLine());
    }
    public static String toStr()throws Exception{
       return br.readLine();
    }
    public static String[] toStrArray()throws Exception{
       String str[]=br.readLine().split(" ");
       return str;
    }
    /****************************************************************/

}
