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.*;

public class Main implements Runnable {
		
	public void solve() throws IOException {
		int N = nextInt();
                
                String a = nextToken();
                String b = nextToken();
                
                int lcm = lcm(a.length(), b.length());
                
                StringBuilder sba = new StringBuilder();
                StringBuilder sbb = new StringBuilder();
                for(int i = 0; i < lcm / a.length(); i++) sba.append(a);
                for(int i = 0; i < lcm / b.length(); i++) sbb.append(b);
                
                int pincha = 0;
                int pinchb = 0;
                //lets play one round
                int tempa = 0;
                int tempb = 0;
                for(int i = 0; i < sba.length(); i++){
                        if(sba.charAt(i) == sbb.charAt(i)) continue;
                        if(sba.charAt(i) == 'R' && sbb.charAt(i) == 'S' || sba.charAt(i) == 'P' && sbb.charAt(i) == 'R' ||sba.charAt(i) == 'S' && sbb.charAt(i) == 'P') tempa++;
                        else if(sbb.charAt(i) == 'R' && sba.charAt(i) == 'S' || sbb.charAt(i) == 'P' && sba.charAt(i) == 'R' ||sbb.charAt(i) == 'S' && sba.charAt(i) == 'P') tempb++;
                }
                
                pincha = (N / lcm) * tempa;
                pinchb = (N/ lcm) * tempb;
                
                N -= lcm * (N / lcm);
                for(int i = 0; i < N; i++){
                        if(sba.charAt(i) == sbb.charAt(i)) continue;
                        if(sba.charAt(i) == 'R' && sbb.charAt(i) == 'S' || sba.charAt(i) == 'P' && sbb.charAt(i) == 'R' ||sba.charAt(i) == 'S' && sbb.charAt(i) == 'P') pincha++;
                        else if(sbb.charAt(i) == 'R' && sba.charAt(i) == 'S' || sbb.charAt(i) == 'P' && sba.charAt(i) == 'R' ||sbb.charAt(i) == 'S' && sba.charAt(i) == 'P') pinchb++;
                }
                
                System.out.println(pinchb + " " + pincha);
	}
        
        private int lcm(int m, int n){
                return (m / gcd(m , n) * n);
        }
        
        private int gcd(int m, int n){
                if(n == 0) return m;
                else return gcd(n, m % n);
        }

	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	//-----------------------------------------------------------
	public static void main(String[] args) {
		new Main().run();
	}

	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			tok = null;
			solve();
			in.close();
		} catch (IOException e) {
			System.exit(0);
		}
	}

	public String nextToken() throws IOException {
		while (tok == null || !tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		return tok.nextToken();
	}

	public int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	public long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	public double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	BufferedReader in;
	StringTokenizer tok;
}