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.awt.Point;
import java.io.*;
import java.math.BigInteger;
import java.util.*;

import static java.lang.Math.*;

public class A implements Runnable{
    
    final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
    
    BufferedReader in;
    PrintWriter out;
    StringTokenizer tok = new StringTokenizer("");
    
    void init() throws FileNotFoundException{
        if (ONLINE_JUDGE){
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        }else{
            in = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter("output.txt");
        }
    }
    
    String readString() throws IOException{
        while(!tok.hasMoreTokens()){
            try{
                tok = new StringTokenizer(in.readLine());
            }catch (Exception e){
                return null;
            }
        }
        return tok.nextToken();
    }
    
    int readInt() throws IOException{
        return Integer.parseInt(readString());
    }
    
    long readLong() throws IOException{
        return Long.parseLong(readString());
    }
    
    double readDouble() throws IOException{
        return Double.parseDouble(readString());
    }
    
    public static void main(String[] args){
        new Thread(null, new A(), "", 256 * (1L << 20)).start();
    }
    
    long timeBegin, timeEnd;

    void time(){
        timeEnd = System.currentTimeMillis();
        System.err.println("Time = " + (timeEnd - timeBegin));
    }
    
    long memoryTotal, memoryFree;
    

    void memory(){
        memoryFree = Runtime.getRuntime().freeMemory();
        System.err.println("Memory = " + ((memoryTotal - memoryFree) >> 10) + " KB");
    }
    
    void debug(Object... objects){
        if (DEBUG){
            for (Object o: objects){
                System.err.println(o.toString());
            }
        }
    }
    
    public void run(){
        try{
            timeBegin = System.currentTimeMillis();
            memoryTotal = Runtime.getRuntime().freeMemory();
            init();
            solve();
            out.close();
            time();
            memory();
        }catch (Exception e){
            e.printStackTrace(System.err);
            System.exit(-1);
        }
    }
    
    boolean DEBUG = false;
    
    void solve() throws IOException{
        long n = readInt();
        
        char[] a = in.readLine().toCharArray();
        int len1 = a.length;
        
        char[] b = in.readLine().toCharArray();
        int len2 = b.length;
        
        long lim = lcm(len1, len2);
        long count1 = 0, count2 = 0;
        
        for (int i = 0, j = 0, k = 0; i < lim; i++, j = (j + 1) % len1, k = (k + 1) % len2){
            if (a[j] == 'R' && b[k] == 'S') count2++;
            if (a[j] == 'S' && b[k] == 'R') count1++;
            
            if (a[j] == 'S' && b[k] == 'P') count2++;
            if (a[j] == 'P' && b[k] == 'S') count1++;
            
            if (a[j] == 'P' && b[k] == 'R') count2++;
            if (a[j] == 'R' && b[k] == 'P') count1++;
        }
        
        count1 *= (n / lim);
        count2 *= (n / lim);
        
        n %= lim;
        for (int i = 0, j = 0, k = 0; i < n; i++, j = (j + 1) % len1, k = (k + 1) % len2){
            if (a[j] == 'R' && b[k] == 'S') count2++;
            if (a[j] == 'S' && b[k] == 'R') count1++;
            
            if (a[j] == 'S' && b[k] == 'P') count2++;
            if (a[j] == 'P' && b[k] == 'S') count1++;
            
            if (a[j] == 'P' && b[k] == 'R') count2++;
            if (a[j] == 'R' && b[k] == 'P') count1++;
        }
        
        out.print(count1 + " " + count2);
    }
    
    long lcm(long a, long b){
        return a / gcd(a, b) * b;
    }
    
    long gcd(long a, long b){
        return (a == 0? b: gcd(b % a, a));
    }
}

