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.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Scanner;

public class Task173A {

	public static void main(String... args) throws NumberFormatException,
			IOException {
		Solution.main(System.in, System.out);
	}

	static class Solution {

		public static void main(InputStream is, OutputStream os)
				throws NumberFormatException, IOException {
			PrintWriter pw = new PrintWriter(os);
			Scanner sc = new Scanner(is);

			int n = sc.nextInt();

			String A = sc.next();
			String B = sc.next();

			int m = A.length();
			int k = B.length();

			{
				StringBuilder sb = new StringBuilder(m * k);
				for (int i = 0; i < k; i++) {
					sb.append(A);
				}
				A = sb.toString();
			}
			{
				StringBuilder sb = new StringBuilder(m * k);
				for (int i = 0; i < m; i++) {
					sb.append(B);
				}
				B = sb.toString();
			}

			if (A.length() != B.length()) {
				pw.flush();
				sc.close();
				throw new RuntimeException("Something went wrong ;-)");
			}

			int retVal1Cycle = 0;
			int retVal1Modulo = 0;
			int retVal2Cycle = 0;
			int retVal2Modulo = 0;

			for (int i = 0; i < m * k; i++) {
				switch (A.charAt(i)) {
				case 'R':
					switch (B.charAt(i)) {
					case 'R':
						break;
					case 'P':
						retVal1Cycle++;
						break;
					case 'S':
						retVal2Cycle++;
						break;
					}
					break;
				case 'P':
					switch (B.charAt(i)) {
					case 'R':
						retVal2Cycle++;
						break;
					case 'P':
						break;
					case 'S':
						retVal1Cycle++;
						break;
					}
					break;
				case 'S':
					switch (B.charAt(i)) {
					case 'R':
						retVal1Cycle++;
						break;
					case 'P':
						retVal2Cycle++;
						break;
					case 'S':
						break;
					}
					break;
				}
			}

			for (int i = 0; i < n % (m * k); i++) {
				switch (A.charAt(i)) {
				case 'R':
					switch (B.charAt(i)) {
					case 'R':
						break;
					case 'P':
						retVal1Modulo++;
						break;
					case 'S':
						retVal2Modulo++;
						break;
					}
					break;
				case 'P':
					switch (B.charAt(i)) {
					case 'R':
						retVal2Modulo++;
						break;
					case 'P':
						break;
					case 'S':
						retVal1Modulo++;
						break;
					}
					break;
				case 'S':
					switch (B.charAt(i)) {
					case 'R':
						retVal1Modulo++;
						break;
					case 'P':
						retVal2Modulo++;
						break;
					case 'S':
						break;
					}
					break;
				}
			}

			int retVal1 = n / (m * k) * retVal1Cycle + retVal1Modulo;
			int retVal2 = n / (m * k) * retVal2Cycle + retVal2Modulo;

			pw.print(retVal1);
			pw.print(" ");
			pw.print(retVal2);

			pw.flush();
			sc.close();
		}
	}

}