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.List;
import java.util.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.util.Arrays;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author codeKNIGHT
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskA solver = new TaskA();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskA {
	public void solve(int testNumber, Scanner in, PrintWriter out) {
        int n=in.nextInt();
        int lcm,gcd,i,q,r;
        String s=in.next(),p=in.next();
        int l1=s.length();
        int l2=p.length();
        gcd= (int)IntegerUtils.gcd(l1,l2);
        lcm=(l1*l2)/gcd;
        q=n/lcm;
        r=n%lcm;
        int c1=0,c2=0,win=0;
        int a[]=new int[lcm];
        for(i=0;i<lcm;i++)
        {
            char ch1,ch2;
            ch1=s.charAt(i%l1);
            ch2=p.charAt(i%l2);
            if(ch1=='R')
            {
                if(ch2=='P')
                   win=2;
                else if(ch2=='S')
                    win=1;
                else win=0;
            }
           else if(ch1=='P')
            {
                if(ch2=='S')
                    win=2;
                else if(ch2=='P')
                    win=0;
                else win=1;
            }
            else
            {
                if(ch2=='S')
                    win=0;
                else if(ch2=='P')
                    win=1;
                else win=2;
            }
            a[i]=win;
            if(win==1)
                c1++;
            else if(win==2)
                c2++;
        }
        c1*=q;
        c2*=q;
        for(i=0;i<r;i++)
        {
            if(a[i]==1)
            {
               c1++;

            }
            else if(a[i]==2)
                c2++;
        }
        out.println(c2+" "+c1);
	}
}

class IntegerUtils {
    public static long gcd(long a,long b)
    {
        if(b==0)
            return a;
        return gcd(b,a%b);
    }

}

