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.StringTokenizer;

public class A {
	BufferedReader in;
	StringTokenizer str;
	PrintWriter out;
	String SK;

	String next() throws IOException {
		while ((str == null) || (!str.hasMoreTokens())) {
			SK = in.readLine();
			if (SK == null)
				return null;
			str = new StringTokenizer(SK);
		}
		return str.nextToken();
	}

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

	double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}

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

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

	void solve() throws IOException {
		int n = nextInt();
		char[] a = next().toCharArray();
		char[] b = next().toCharArray();
		int t1 = a.length;
		int t2 = b.length;
		int nok = t1 * t2 / gcd(t1, t2);
		int save = n % nok;
		int r1 = 0;
		int r2 = 0;
		int[] win = new int[3];
		for (int i = 0; i < nok; i++) {
			int tek = 2;
			int pi = i % t1, pj = i % t2;
			if (a[pi] != b[pj]) {
				if (a[pi] == 'R' && b[pj] == 'S' || a[pi] == 'S'
						&& b[pj] == 'P' || a[pi] == 'P' && b[pj] == 'R') {
					tek = 0;
				} else
					tek = 1;
			}
			win[tek]++;
			if (i + 1 == save) {
				r1 += win[0];
				r2 += win[1];
			}
		}
		out.println(win[1] * (n / nok) + r2 + " " + (win[0] * (n / nok) + r1));
	}

	void run() throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

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

}