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.util.Scanner;

// 173A
public class RockPaperScissors {

	public static void main(String[] args) { 
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		String a = in.next();
		String b = in.next();
		
		int lcm = lcm(a.length(), b.length());
		
		StringBuilder sba = new StringBuilder(a);
		for (int i = 0; sba.length() < lcm; i = (i + 1) % a.length()) {
			sba.append(a.charAt(i));
		}
		a = sba.toString();
		
		StringBuilder sbb = new StringBuilder(b);
		for (int i = 0; sbb.length() < lcm; i = (i + 1) % b.length()) {
			sbb.append(b.charAt(i));
		}
		b = sbb.toString();
				
		int aLosses = 0;
		int bLosses = 0;
		for (int i = 0; i < Math.min(n, a.length()); i++) {
			if (a.charAt(i) == 'R' && b.charAt(i) == 'S' ||
				a.charAt(i) == 'P' && b.charAt(i) == 'R' ||
				a.charAt(i) == 'S' && b.charAt(i) == 'P')
				bLosses++;
			else if (a.charAt(i) != b.charAt(i))
				aLosses++;
		}
		
		aLosses = aLosses * (n / a.length());
		bLosses = bLosses * (n / b.length());
		
		for (int i = 0; i < n % a.length(); i++) {
			if (a.charAt(i) == 'R' && b.charAt(i) == 'S' ||
					 a.charAt(i) == 'P' && b.charAt(i) == 'R' ||
					 a.charAt(i) == 'S' && b.charAt(i) == 'P')
				bLosses++;
			else if (a.charAt(i) != b.charAt(i))
				aLosses++;
		}
		
		System.out.printf("%d %d\n", aLosses, bLosses);		
	}
	static int lcm(int a, int b) {
		return b * a / gcd(a, b);
	}
	static int gcd(int a, int b) {
		if (b == 0)
			return a;
		return gcd(b, a % b);
	}
}
