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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;

public class Test {
	static BufferedReader reader;
	static StringTokenizer tokenizer;
	static PrintWriter writer;

	static int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	static long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	static double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	static String nextToken() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			tokenizer = new StringTokenizer(reader.readLine());
		}
		return tokenizer.nextToken();
	}

	public static void main(String[] args) throws IOException {
		reader = new BufferedReader(new InputStreamReader(System.in));
		tokenizer = null;
		writer = new PrintWriter(System.out);
		solve();
		reader.close();
		writer.close();
	}

	static int gcd (int a, int b) {
		if (b == 0)
			return a;
		else
			return gcd (b, a % b);
	}
	
	private static void solve() throws IOException {
		int n = nextInt();
		String a = nextToken();
		String b = nextToken();
		
		int m = a.length();
		int k = b.length();
		
		int g = m*k/gcd(m,k);
		
		
		
		int sa=0, sb=0;
		for(int i=0;i<g;i++)
		{
			int c1=i%m;
			int c2=i%k;
			if(a.charAt(c1)=='R'&&b.charAt(c2)=='P')
			{
				sa++;
			}
			else if(a.charAt(c1)=='R'&&b.charAt(c2)=='S')
			{
				sb++;
			}
			else if(a.charAt(c1)=='P'&&b.charAt(c2)=='R')
			{
				sb++;
			}
			else if(a.charAt(c1)=='P'&&b.charAt(c2)=='S')
			{
				sa++;
			}
			else if(a.charAt(c1)=='S'&&b.charAt(c2)=='P')
			{
				sb++;
			}
			else if(a.charAt(c1)=='S'&&b.charAt(c2)=='R')
			{
				sa++;
			}
		}
		
		int s1=0, s2=0;
		
		s1+=(n/g)*sa;
		s2+=(n/g)*sb;
		
		int n1=n%g;
		
		for(int i=0;i<n1;i++)
		{
			int c1=i%m;
			int c2=i%k;
			if(a.charAt(c1)=='R'&&b.charAt(c2)=='P')
			{
				s1++;
			}
			else if(a.charAt(c1)=='R'&&b.charAt(c2)=='S')
			{
				s2++;
			}
			else if(a.charAt(c1)=='P'&&b.charAt(c2)=='R')
			{
				s2++;
			}
			else if(a.charAt(c1)=='P'&&b.charAt(c2)=='S')
			{
				s1++;
			}
			else if(a.charAt(c1)=='S'&&b.charAt(c2)=='P')
			{
				s2++;
			}
			else if(a.charAt(c1)=='S'&&b.charAt(c2)=='R')
			{
				s1++;
			}
		}
		
		writer.println(s1+" "+s2);
	}
}