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.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Date;
import java.util.Timer;
import java.util.TimerTask;
import java.util.TreeMap;
import java.util.TreeSet;

public class Main{
	
	private static int gcd (int a, int b) {
		return b > 0 ? gcd (b, a % b) : a;
	}
	
	private static int lcm (int a, int b){
		return a / gcd (a, b) * b;
	}
	
	public static void main(String[] args) throws Exception {
		
		int n = Integer.parseInt(inB.readLine());
		char[] tmpnik = inB.readLine().toCharArray();
		char[] tmppol = inB.readLine().toCharArray();
		int lcm = lcm(tmpnik.length, tmppol.length);
		char[] nik = new char[lcm];
		char[] pol = new char[lcm];
		int[] winner = new int[lcm]; 
		int niks = 0;
		int pols = 0;
		for(int i = 0; i < lcm; i++) nik[i] = tmpnik[i % (tmpnik.length)];
		for(int i = 0; i < lcm; i++){
			pol[i] = tmppol[i % (tmppol.length)];
			if(pol[i] == nik[i]) continue;
			else if(pol[i] == 'R' && nik[i] == 'S') winner[i] = -1;
			else if(pol[i] == 'R' && nik[i] == 'P') winner[i] = 1;
			else if(pol[i] == 'S' && nik[i] == 'R') winner[i] = 1;
			else if(pol[i] == 'S' && nik[i] == 'P') winner[i] = -1;
			else if(pol[i] == 'P' && nik[i] == 'R') winner[i] = -1;
			else if(pol[i] == 'P' && nik[i] == 'S') winner[i] = 1;
			if(winner[i] == 1) pols++;
			else if(winner[i] == -1) niks++;
		}
		int anspols = pols * (n / lcm);
		int ansniks = niks * (n / lcm);
		int nn = n % lcm;
		for(int i = 0; i < nn; i++){
			if(winner[i] == 1) anspols++;
			else if(winner[i] == -1) ansniks++;
		}
		println(ansniks + " " + anspols);
	}
		
	private static PrintWriter out;
	private static BufferedReader inB;
	
	static {
		try {
			out = new PrintWriter(System.out);
			inB = new BufferedReader(new InputStreamReader(System.in));
		} catch(Exception e) {e.printStackTrace();}
	}
	
	private static StreamTokenizer in = new StreamTokenizer(inB);
	
	private static void exit(Object o) throws Exception {
		out.println(o);
		out.flush();
		System.exit(0);
	}
	private static void println(Object o) throws Exception{
		out.println(o);
		out.flush();
	}
	private static void print(Object o) throws Exception{
		out.print(o);
		out.flush();
	}
	private static int nextInt() throws Exception {
		in.nextToken();
		return (int)in.nval;
	}  
	private static String nextString() throws Exception {
		in.nextToken();
		return in.sval;        
	}
} 