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.*;
import java.util.*;

public class cf173A {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
      Scanner in =new Scanner(System.in);
      int n=in.nextInt();
      
      String s1=in.next();
      String s2=in.next();
      
      int cl=(s1.length()*s2.length())/gcd(s1.length(),s2.length());
      //int g=gcd(s1.length(),s2.length());
      //System.out.println(cl);
      
      int arr[][]=new int[cl][2];
      
     if(true) {
      char c1=s1.charAt(0%(s1.length()));
	  char c2=s2.charAt(0%(s2.length()));
	  
	  if((c1=='R' && c2=='S') || (c1=='S' && c2=='P') || (c1=='P' && c2=='R') )
		  arr[0][1]=1;
	  
	  if((c2=='R' && c1=='S') || (c2=='S' && c1=='P') || (c2=='P' && c1=='R') )
		  arr[0][0]=+1;
     }
      
      for(int i=1;i<cl;i++) {
    	  char c1=s1.charAt(i%(s1.length()));
    	  char c2=s2.charAt(i%(s2.length()));
    	  
    	  if((c1=='R' && c2=='S') || (c1=='S' && c2=='P') || (c1=='P' && c2=='R') )
    	  {  arr[i][1]=arr[i-1][1]+1;
    	  	  arr[i][0]=arr[i-1][0];
    	  }
    	  
    	  else if((c2=='R' && c1=='S') || (c2=='S' && c1=='P') || (c2=='P' && c1=='R') )
    	  {  arr[i][0]=arr[i-1][0]+1;
    	     arr[i][1]=arr[i-1][1];
    	  }
    	  else
    	  {
    		  arr[i][1]=arr[i-1][1];
    		  arr[i][0]=arr[i-1][0];
    	  }
    	  
    	  
      }
      
      long ans1,ans2;
      int div=n/cl;
      int rem=n%cl;
      if(rem==0) {
    	   ans1=(arr[cl-1][0] * div) ;
           ans2=(arr[cl-1][1] * div) ;
    	  
      }
      else {
    	   ans1=(arr[cl-1][0] * div) + arr[rem-1][0];
           ans2=(arr[cl-1][1] * div) + arr[rem-1][1];
    	  
      }
      
      
      System.out.println(ans1+" "+ans2);
      
      /*for(int i=0;i<cl;i++) {
    	  System.out.println(arr[i][0]+" "+arr[i][1]);
      } */
	}
	
	static int gcd(int a,int b)
	{
		if(b>a)
		{
			int temp=a;
			a=b;b=temp;
		}
		
		if(a%b ==0) return b;
		else return gcd(b,a%b);
			
	}
	

}
