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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Collections;
import java.util.Locale;
import java.util.StringTokenizer;

public class A {

	private int gcd(int a, int b) {
		if (a == 0) {
			return b;
		} else {
			return gcd(b % a, a);
		}
	}
	
	private int lcm(int a, int b) {
		return a * b / gcd(a, b);
	}
	
	private void solve() throws IOException {
		long n = nextInt();
		String s1 = nextToken();
		String s2 = nextToken();
		
		int l = lcm(s1.length(), s2.length());
		char a1[] = new char[l];
		char a2[] = new char[l];
		for (int i = 0; i < l; i++) {
			a1[i] = s1.charAt(i % s1.length());
			a2[i] = s2.charAt(i % s2.length());
		}
		
		int[] w1 = new int[l];
		int[] w2 = new int[l];
		if (a1[0] == 'R' && a2[0] == 'S' || a1[0] == 'S' && a2[0] == 'P' || a1[0] == 'P' && a2[0] == 'R') {
			w1[0] = 1;
		} else if (a1[0] != a2[0]) {
			w2[0] = 1;
		}
		
		for (int i = 1; i < l; i++) {
			if (a1[i] == 'R' && a2[i] == 'S' || a1[i] == 'S' && a2[i] == 'P' || a1[i] == 'P' && a2[i] == 'R') {
				w1[i] = w1[i - 1] + 1;
				w2[i] = w2[i - 1];
			} else if (a1[i] != a2[i]){
				w2[i] = w2[i - 1] + 1;
				w1[i] = w1[i - 1];
			} else {
				w1[i] = w1[i - 1];
				w2[i] = w2[i - 1];
			}
		}
		
		long A = n / l * w1[l - 1];
		long B = n / l * w2[l - 1];
		if (n % l != 0) {
			A += w1[(int) (n % l - 1)];
			B += w2[(int) (n % l - 1)];
		}
		
		println(B + " " + A);
	}
	
	private String nextToken() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			tokenizer = new StringTokenizer(reader.readLine());
		}
		return tokenizer.nextToken();
	}

	private int nextInt() throws NumberFormatException, IOException {
		return Integer.parseInt(nextToken());
	}

	private double nextDouble() throws NumberFormatException, IOException {
		return Double.parseDouble(nextToken());
	}

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

	private void print(Object o) {
		writer.print(o);
	}

	private void println(Object o) {
		writer.println(o);
	}

	private void printf(String format, Object... o) {
		writer.printf(format, o);
	}

	public static void main(String[] args) {
		long time = System.currentTimeMillis();
		Locale.setDefault(Locale.US);
		new A().run();
		System.err.printf("%.3f\n", 1e-3 * (System.currentTimeMillis() - time));
	}

	BufferedReader reader;
	StringTokenizer tokenizer;
	PrintWriter writer;

	private void run() {
		try {
			reader = new BufferedReader(new InputStreamReader(System.in));
			writer = new PrintWriter(System.out);
			solve();
			reader.close();
			writer.close();
		} catch (IOException e) {
			e.printStackTrace();
			System.exit(13);
		}
	}
}