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.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Stack;
import java.util.StringTokenizer;

public class Main {

    FastScanner in = new FastScanner(System.in);
//    FastScanner in = new FastScanner(new File("test.txt"));
    
    PrintWriter out = new PrintWriter(System.out);

    public static void main (String[]args) throws IOException {
        Main task = new Main();
        task.solve();
        task.close();
    }

    public void close () {
        in.close();
        out.close();
    }
    
    public void solve() throws IOException {
        int n = in.nextInt();
        String a = in.next();
        String b = in.next();
        
        int AA = 0;
        int BB = 0;
        
        int Len = Algebra.lcm(a.length(), b.length());
        StringBuilder sb = new StringBuilder();
        while (sb.length() != Len) {
            sb.append(a);
        }
        a = sb.toString();
        
        sb = new StringBuilder();
        while (sb.length() != Len) {
            sb.append(b);
        }
        b = sb.toString();
        
        int[]ans = new int[Len];
        for (int i = 0; i < a.length(); i++) {
            char aa = a.charAt(i);
            char bb = b.charAt(i);
            if (aa == bb) ans[i] = 0;
            if (aa == 'S' && bb == 'P') ans[i] = 1;
            if (aa == 'S' && bb == 'R') ans[i] = 2;
            if (aa == 'P' && bb == 'S') ans[i] = 2;
            if (aa == 'P' && bb == 'R') ans[i] = 1;
            if (aa == 'R' && bb == 'P') ans[i] = 2;
            if (aa == 'R' && bb == 'S') ans[i] = 1;
        }
        
        int div = n / Len;
        int ost = n % Len;
        
        int acount = 0;
        int bcount = 0;
        for (int i = 0; i < Len; i++) {
            if (ans[i] == 1) acount++;
            if (ans[i] == 2) bcount++;
        }
        
        AA += div * acount;
        BB += div * bcount;
        
        for (int i = 0; i < ost; i++) {
            if (ans[i] == 1) {
                AA++;
            }
            if (ans[i] == 2) {
                BB++;
            }
        }
        System.out.println(BB + " " + AA);
    }      
}


class Algebra {
    /****
     *       Number of co-prime numbers on [1, n].
     *       Number a is Co-prime if gcd (a, n) == 1
     *       O (sqrt(n))
     ****/
    public static int phi(int n) {
        int result = n;
        for (int i = 2; i*i <= n; ++i) {
            if (n % i == 0) {
                while (n % i == 0) {
                    n /= i;
                }
                result -= result / i;
            }
        }
        if (n > 1) {
            result -= result / n;
        }
        return result;
    }

    /****
     *       Raise number a to power of n.
     *       O (log n)
     ****/
    public static int binpow (long a, int n) {
        int res = 1;
        while (n != 0) {
            if ((n & 1) == 1)
                res *= a;
            a *= a;
            n >>= 1;
        }
        return res;
    }

    /****
     *       Finding the greatest common divisor of two numbers.
     *       O (log min(a, b))
     ****/
    public static int gcd (int a, int b) {
        return (b != 0) ? gcd (b, a % b) : a;
    }

    /****
     *       Finding the lowest common multiple of two numbers.
     *       O (log min(a, b))
     ****/
    public static int lcm (int a, int b) {
        return a / gcd (a, b) * b;
    }

    /****
     *       Eratosthenes Sieve of numbers - [0..n]. True - simple, False - not simple.
     *       O (n log log n)
     ****/
    public static boolean[] sieveOfEratosthenes (int n) {
        boolean [] prime = new boolean[n + 1];
        Arrays.fill(prime, true);
        prime[0] = prime[1] = false;
        for (int i=2; i<=n; ++i) {
            if (prime[i]) {
                if (i * 1L * i <= n) {
                    for (int j=i*i; j<=n; j+=i) {
                        prime[j] = false;
                    }
                }
            }
        }
        return prime;
    }    
}

class FastScanner {
    BufferedReader br;
    StringTokenizer st;
    
    FastScanner(File f) {
        try {
            br = new BufferedReader(new FileReader(f));
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }
    }
    
    FastScanner(InputStream in) {
        br = new BufferedReader(new InputStreamReader(in));
    }
    
    String next() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                System.err.println(e);
                return "";
            }
        }
        return st.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    long nextLong() {
        return Long.parseLong(next());
    }

    double nextDouble() {
        return Double.parseDouble(next());
    }

    float nextFloat() {
        return Float.parseFloat(next());
    }
    BigInteger nextBigInt() {
        return new BigInteger(next());
    }

    void close() {
        try {
            br.close();
        }
        catch (IOException e) {
        }
    }
}