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.*;
import java.util.*;

public class Solution {
	public static StringTokenizer tok = null;
	public static BufferedReader cin;
	
	public static String nextToken() throws Exception {
		if (tok == null || !tok.hasMoreTokens()) {
			tok = new StringTokenizer(cin.readLine());
		}
		return tok.nextToken();
	}
	
	public static int nextInt() throws Exception {
		return Integer.parseInt(nextToken());
	}
	
	public static double nextDouble() throws Exception {
		return Double.parseDouble(nextToken());
	}
	
	public static void main(String[] args) throws Exception {
		cin = new BufferedReader(new InputStreamReader(System.in));
		int n = nextInt();
		String A = nextToken();
		String B = nextToken();
		int alen = A.length();
		int blen = B.length();
		int m = alen * blen;
		int alose = 0;
		int blose = 0;
		for (int i = 0; i < m; i++) {
			char ach = A.charAt(i % alen);
			char bch = B.charAt(i % blen);
			if (ach != bch) {
				if (ach == 'R') {
					if (bch == 'P') {
						alose++;
					} else {
						blose++;
					}
				} else
				if (ach == 'P') {
					if (bch == 'S') {
						alose++;
					} else {
						blose++;
					}
				} else
				if (ach == 'S') {
					if (bch == 'R') {
						alose++;
					} else {
						blose++;
					}
				}
			}
		}
		int cnt = n / m;
		int k = n % m;
		alose *= cnt;
		blose *= cnt;
		for (int i = 0; i < k; i++) {
			char ach = A.charAt(i % alen);
			char bch = B.charAt(i % blen);
			if (ach != bch) {
				if (ach == 'R') {
					if (bch == 'P') {
						alose++;
					} else {
						blose++;
					}
				} else
				if (ach == 'P') {
					if (bch == 'S') {
						alose++;
					} else {
						blose++;
					}
				} else
				if (ach == 'S') {
					if (bch == 'R') {
						alose++;
					} else {
						blose++;
					}
				}
			}
		}
		System.out.println(alose + " " + blose);
	}

}
