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.*;

public class a {
	public static void main(String[] args) {
		new a();
	}

	public a() {
		Scanner in = new Scanner(System.in);

		long games = in.nextInt();
		String a = in.next();
		String b = in.next();

		int lcm = lcm(a.length(), b.length());

		StringBuilder sb = new StringBuilder();
		for(int i=0; i<lcm/a.length(); ++i) sb.append(a);
		a = sb.toString();

		sb = new StringBuilder();
		for(int i=0; i<lcm/b.length(); ++i) sb.append(b);
		b = sb.toString();

		// score[0] = player A, sore[1] = player B
		int[] score = new int[2];

		long pos = 0;

		if(games >= lcm) {
			// Get score up to lcm
			for(int i=0; i<lcm; ++i) {
				update(score, a.charAt(i), b.charAt(i));
			}

			int numCopies = (int)(games/lcm);
			score[0] *= numCopies;
			score[1] *= numCopies;

			pos = numCopies*lcm;
		}

		for(; pos<games; ++pos) {
			// System.out.println("Here " + a.charAt((int)(pos%lcm)) + " " + b.charAt((int)(pos%lcm)));
			update(score, a.charAt((int)(pos%lcm)), b.charAt((int)(pos%lcm)));
		}
		// System.out.println(a.substring(0, (int)games) + "\n" + b.substring(0, (int)games) + " \n");
		System.out.println(score[1] + " " + score[0]);
	}

	private void update(int[] score, char a, char b) {
		if(a == 'P' && b == 'R') ++score[0];
		else if(a == 'R' && b == 'P') ++score[1];

		else if(a == 'S' && b == 'P') ++score[0];
		else if(a == 'P' && b == 'S') ++score[1];

		else if(a == 'R' && b == 'S') ++score[0];
		else if(a == 'S' && b == 'R') ++score[1];
	}

	private int lcm(int a, int b) {
		return (a/gcd(a, b))*b;
	}

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