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 A {

	final String filename = new String("A").toLowerCase();

	int get(char a, char b) {
		if (a == b) {
			return 0;
		}
		if ((a == 'R' && b == 'S') || (a == 'S' && b == 'P')
				|| (a == 'P' && b == 'R')) {
			return 1;
		} else {
			return 2;
		}
	}

	int gcd(int a, int b) {
		if (a == 0) {
			return b;
		}
		return gcd(b % a, a);
	}

	void solve() throws Exception {
		int n = nextInt();
		String s1 = nextToken();
		String s2 = nextToken();
		int lcm = s1.length() * s2.length() / gcd(s1.length(), s2.length());
		int[] win = new int[lcm];
		int sum1 = 0;
		int sum2 = 0;
		for (int i = 0; i < lcm; i++) {
			int x = i % s1.length();
			int y = i % s2.length();
			win[i] = get(s1.charAt(x), s2.charAt(y));
			if (win[i] == 1) {
				sum1++;
			}
			if (win[i] == 2) {
				sum2++;
			}
		}
		
		int count = n / lcm;
		sum1 *= count;
		sum2 *= count;
		for (int i = 0; i < n % lcm; i++) {
			if (win[i] == 1) {
				sum1++;
			}
			if (win[i] == 2) {
				sum2++;
			}
		}
		out.println(sum2 + " " + sum1);
	}

	void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			// in = new BufferedReader(new FileReader("input.txt"));
			// out = new PrintWriter("output.txt");

			solve();

			out.close();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

	BufferedReader in;
	StringTokenizer st;
	PrintWriter out;

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

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

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

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

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

}
