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 RPS
{
	public static void main(String[] args) throws IOException
	{
	 	InputStreamReader isr = new InputStreamReader(System.in);
		BufferedReader br = new BufferedReader(isr);
		
		int n = Integer.parseInt(br.readLine());
		String N = br.readLine();
		String P = br.readLine();
		int l1 = N.length();
		int l2 = P.length();

		int l3 = gcd(l1,l2);
//		System.out.println(l1+" "+l2+" "+l3);
		l3 = (l2*l1)/l3;		
		char[] res = new char[l3];

		for(int i=0;i<l3;i++)
		{
			int a=i%l1;
			int b=i%l2;

			if(N.charAt(a) == 'R')
			{
				if(P.charAt(b) == 'S')
					res[i] = 'N';
				else if(P.charAt(b) == 'P')
					res[i] = 'P';
				else 
					res[i] = 'D';
				
			}	

			if(N.charAt(a) == 'S')
			{
				if(P.charAt(b) == 'S')
					res[i] = 'D';
				else if(P.charAt(b) == 'P')
					res[i] = 'N';
				else 
					res[i] = 'P';
			}	
			if(N.charAt(a) == 'P')
			{
				if(P.charAt(b) == 'S')
					res[i] = 'P';
				else if(P.charAt(b) == 'P')
					res[i] = 'D';
				else 
					res[i] = 'N';
			}	
		}

		int nr,pr;

		nr = pr =0;

		for(int i=0;i<l3;i++)
		{		
			if(res[i] == 'N')
				pr++;
			else if(res[i] == 'P')
				nr++;
		}
		
		int q = n/l3;

		nr = nr*q;
		pr = pr*q;

		int r = n%l3;

		for(int i=0;i<r;i++)
		{		
			if(res[i] == 'N')
				pr++;
			else if(res[i] == 'P')
				nr++;
		}

		System.out.println(nr+" "+pr);  
	}

	public static int gcd(int a, int b)
	{
//		System.out.println(a+" "+b);
		return (b==0) ? a : gcd(b,a%b);
	}
}
