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.*;
import java.io.*;

public class Fuck {

	BufferedReader br;
	StringTokenizer st;
	PrintWriter out;
	
	public static void main(String[] args) throws IOException {
		new Fuck().run();
	}
	
	int gcd(int x, int y) {
		if (y == 0)
			return x;
		else return gcd(y, x % y);
	}
	
	void run() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		
		int n = nextInt();
		char[] as = next().toCharArray();
		char[] bs = next().toCharArray();
		
		//System.err.println(Arrays.toString(as));
		//System.err.println(Arrays.toString(bs));
		
		int a = as.length;
		int b = bs.length;
		
		int lcm = (a / gcd(a, b)) * b;
		int count1 = 0, count2 = 0;
		
		//System.err.println(lcm);
		
		for (int i = 0; i < lcm; i++) {
			char c1 = as[i % as.length];
			char c2 = bs[i % bs.length];
			
			if (c1 == c2)
				continue;
			else if (c1 == 'R' && c2 == 'S' || c1 == 'P' && c2 == 'R' || c1 == 'S' && c2 == 'P')
				count1++;
			else
				count2++;
		}
		
		int ans1 = count1 * (n / lcm);
		int ans2 = count2 * (n / lcm);
		
		n = n % lcm;
		for (int i = 0; i < n; i++) {
			char c1 = as[i % as.length];
			char c2 = bs[i % bs.length];
			
			if (c1 == c2)
				continue;
			else if (c1 == 'R' && c2 == 'S' || c1 == 'P' && c2 == 'R' || c1 == 'S' && c2 == 'P')
				ans1++;
			else
				ans2++;
		}
		
		out.printf("%d %d\n", ans2, ans1);
		
		out.close();
	}
	
	String next() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(br.readLine());
		}
		return st.nextToken();
	}
	
	int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
	
}
