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.*;

public class Solution {


	private StringTokenizer st;
	private BufferedReader in;
	private PrintWriter out;

	public void solve() throws IOException {
		int n = nextInt();
		String u = next();
		String v = next();
		int ans1 = 0;
		int ans2 = 0;
		final String s = "RSP";
		int m = u.length() * v.length();
		for (int it = 0, i = 0, j = 0; it < m; ++it, i++, j++) {
			if (i == u.length()) {
				i = 0;
			}
			if (j == v.length()) {
				j = 0;
			}
			int val = (n + m - it - 1) / m;
			int a = s.indexOf(u.charAt(i));
			int b = s.indexOf(v.charAt(j));
			if ((a + 1) % 3 == b) {
				ans2 += val;
			}
			if ((b + 1) % 3 == a) {
				ans1 += val;
			}
		}
		out.println(ans1 + " " + ans2);
	}

	public void run() throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		eat("");
		
		solve();
		
		out.close();
	}
	
	void eat(String s) {
		st = new StringTokenizer(s);
	}
	
	String next() throws IOException {
		while (!st.hasMoreTokens()) {
			String line = in.readLine();
			if (line == null) {
				return null;
			}
			eat(line);
		}
		return st.nextToken();
	}
	
	int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
	
	long nextLong() throws IOException {
		return Long.parseLong(next());
	}
	
	double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
	
	public static void main(String[] args) throws IOException {
		Locale.setDefault(Locale.US);
		new Solution().run();
	}
	
}