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 Shishki implements Runnable {
	public static void main(String[] args) {
		new Thread(new Shishki()).run();
	}

	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer in;
	PrintWriter out = new PrintWriter(System.out);

	public String nextToken() throws IOException {
		while (in == null || !in.hasMoreTokens()) {
			in = new StringTokenizer(br.readLine());
		}

		return in.nextToken();
	}

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

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

	public void solve() throws IOException {
		int n = nextInt();

		char[] mz = nextToken().toCharArray();
		char[] kz = nextToken().toCharArray();

		int m = mz.length;
		int k = kz.length;

		int[] ma = new int[m * k];
		int[] ka = new int[m * k];

		for (int i = 0; i < k; i++) {
			for (int j = 0; j < m; j++) {
				ma[i * m + j] = mz[j] - 'P';
				if (ma[i * m + j] > 0)
					ma[i * m + j]--;
			}
		}

		for (int i = 0; i < m; i++) {
			for (int j = 0; j < k; j++) {
				ka[i * k + j] = kz[j] - 'P';
				if (ka[i * k + j] > 0)
					ka[i * k + j]--;
			}
		}

		int[][] count = new int[2][m * k + 1];

		for (int i = 0; i < m * k; i++) {
			count[0][i + 1] = count[0][i];
			count[1][i + 1] = count[1][i];

			if ((ma[i] + 1) % 3 == ka[i]) {
				count[1][i + 1]++;
			}
			if ((ka[i] + 1) % 3 == ma[i]) {
				count[0][i + 1]++;
			}
		}

		int l = m * k;
		int[] ans = new int[2];
		ans[0] = count[0][l] * (n / l) + count[0][n % l];
		ans[1] = count[1][l] * (n / l) + count[1][n % l];

		out.println(ans[0] + " " + ans[1]);
	}

	public void run() {
		try {
			solve();
			out.close();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}
}
