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.InputStreamReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;

public class palin {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(System.out);
		Scanner scan = new Scanner(System.in);
		TaskC solver = new TaskC();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskC {
	public void solve(int testNumber, InputReader in, PrintWriter out) {
		int n = in.nextInt(), pa = 0, pb = 0, d;
		String a = in.next();
		String b = in.next();
		for (int i = 0; i < n && i < a.length() * b.length(); i++) {
			d = (n - i + a.length() * b.length() - 1)
					/ (a.length() * b.length());
			char ca = a.charAt(i % a.length());
			char cb = b.charAt(i % b.length());
			if (ca == 'R' && cb == 'S' || ca == 'S' && cb == 'P' || ca == 'P'
					&& cb == 'R') {
				pb += d;
			} else {
				if (ca != cb) {
					pa += d;
				}
			}
		}
		out.print(pa + " " + pb);
	}
}

class InputReader {
	BufferedReader br;
	StringTokenizer st;

	public InputReader(InputStream in) {
		br = new BufferedReader(new InputStreamReader(in));
		st = null;
	}

	public String next() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return st.nextToken();
	}

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

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