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.Arrays;
import java.util.Scanner;

public class crocb1 {

    public static void debug(Object... obs)
    {
        System.out.println(Arrays.deepToString(obs));
    }

    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        
        int n=sc.nextInt();
        String ls=sc.next();
        String ms=sc.next();
        int l=ls.length();
        int m=ms.length();
        
        int lcm=lcm(l,m);
        
        int r = n%lcm;
        int pp = 0;
        int mp = 0;
        
        int ppr = 0;
        int mpr = 0;
        
        for(int i=0;i<lcm;i++)
        {
            if(i==r)
            {
                ppr = pp;
                mpr = mp;
            }
            
            if(beats(ls.charAt(i%l),ms.charAt(i%m)))
            {
                pp++;
            }
            else if(beats(ms.charAt(i%m),ls.charAt(i%l)))
            {
                mp++;
            }
        }
        
        //debug(lcm,n,n/lcm,r,pp,ppr,mp,mpr);
        pp=(n/lcm) * pp + ppr;
        mp=(n/lcm) * mp + mpr;
        
        
        System.out.println(mp + " " + pp);
    }

    private static boolean beats(char c1, char c2)
    {
        if(c1 == 'R' && c2 == 'S')
            return true;
        if(c1 == 'S' && c2 == 'P')
            return true;
        if(c1 == 'P' && c2 == 'R')
            return true;
        return false;
    }

    private static int lcm(int l, int m)
    {
        return l * (m/gcd(m,l));
    }

    private static int gcd(int m, int l)
    {
        while(l!=0)
        {
            int tmp = m%l;
            m = l;
            l = tmp;
        }
        return m;
    }
}