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 RockPaperScissors {
	
	static int[] A, B;

	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(in.readLine());
		A = toArray(in.readLine());
		B = toArray(in.readLine());
		int[] res = solve(n);
		System.out.println(res[0]+" "+ res[1]);
	}
	
	static int[] solve(int n) {
		int l = lcm(A.length, B.length);
		int[] res = {0, 0};
		if (n <= l) {
			for (int i = 0; i < n; i++) {
				play(res, A[i%A.length], B[i%B.length]);
			}
			return res;
		} else {
			int[] period = solve(l);
			int m = n / l;
			for (int i = 0; i < 2; i++) {
				res[i] = m*period[i];
			}
			int[] rest = solve(n%l);
			for (int i = 0; i < 2; i++) {
				res[i] += rest[i];
			}
			return res;
		}
	}
	
	static void play(int[] res, int a, int b) {
		if (b == a + 1 || b == 0 && a == 2) {
			res[0]++;
		} else if (a == b + 1 || a == 0 && b == 2) {
			res[1]++;
		}
	}
	
	static int lcm(int a, int b) {
		return a*b/gcd(a, b);
	}
	
	static int gcd(int a, int b) {
		return b == 0 ? a : gcd(b, a%b);
	}
	
	static int[] toArray(String s) {
		int[] res = new int[s.length()];
		for (int i = 0; i < s.length(); i++) {
			char c = s.charAt(i);
			if (c == 'R') {
				res[i] = 0;
			} else if (c == 'P') {
				res[i] = 1;
			} else {
				res[i] = 2;
			}
		}
		return res;
	}
}
