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.lang.*;
import java.io.*;
public class Main{

    static long gcdEuclid(long a,long b) {
		if(b==0) return a;
		else return gcdEuclid(b,a%b);
	}
	static long getlcm(long a,long b) {
		long gcd = gcdEuclid(a,b);
		return (a*b)/gcd;
	}
	
    
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
       
       int n = sc.nextInt();
       String ss1 = sc.next();
       String ss2 = sc.next();
       
      StringBuilder s1 = new StringBuilder();
      s1.append(ss1);
      StringBuilder s2 = new StringBuilder();
      s2.append(ss2);
       
       int l1 = s1.length();
       int l2 = s2.length();
       
       char cc1 = s1.charAt(0);
        char cc2 = s2.charAt(0);
        int ra = 0;
       int rb = 0;
           
       int lcm = (int)getlcm(l1,l2);
       
       
       for(int i=0;i<lcm;i++){
           
           cc1 = s1.charAt(i%l1);
            cc2 = s2.charAt(i%l2);
           if(cc1!=cc2){
            if((cc1=='R' && cc2=='S') ||(cc1=='S' && cc2=='P') || (cc1=='P' && cc2=='R') )
                 ra++;
           else rb++;
           }
           
       }
       int nnn = n/lcm;
       ra = nnn*ra;
       rb = nnn*rb;
       
       for(int i=0;i<n%lcm;i++){
           cc1 = s1.charAt(i%l1);
            cc2 = s2.charAt(i%l2);
           if(cc1!=cc2){
            if((cc1=='R' && cc2=='S') ||(cc1=='S' && cc2=='P') || (cc1=='P' && cc2=='R') )
                 ra++;
           else rb++;
           }
       }
       
       System.out.println(rb+ " " + ra);
       
       
       
    }
}