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 rps {

	public static void main(String [] args)throws Exception{
		new rps().input();
	}
	public void input() throws Exception{
		BufferedReader bb= new BufferedReader(new InputStreamReader(System.in));
		int n=Integer.parseInt(bb.readLine());
		char [] a=bb.readLine().toCharArray();
		char [] b=bb.readLine().toCharArray();
		solve(a,b,n);
	}
	
	public void solve(char [] a, char [] b, int n){
		int g=gcd(Math.max(a.length, b.length),Math.min(a.length, b.length));
		int lcm=a.length*b.length/g;
		int [] t=new int[2];
		if(lcm>=n){
			char [] a1=new char[n];
			char [] b1= new char[n];
			int i=0;
			while(i<n){
				a1[i]=a[i%a.length];
				b1[i]=b[i%b.length];
				i++;
			}
			t=compare(a1,b1);
			
		}
		else{
			char [] a1=new char[lcm];
			char [] b1= new char[lcm];
			int i=0;
			while(i<lcm){
				a1[i]=a[i%a.length];
				b1[i]=b[i%b.length];
				i++;
			}
			 t=compare(a1,b1);
			t[0]=t[0]*(n/lcm);
			t[1]=t[1]*(n/lcm);
			a1=null;
			b1=null;
			a1= new char[n%lcm];
			b1=new char[n%lcm];
			i=0;
			while(i<n%lcm){
				a1[i]=a[i%a.length];
				b1[i]=b[i%b.length];
				i++;
			}
			int [] t1=compare(a1,b1);
			t[0]+=t1[0];
			t[1]+=t1[1];
			
		}
		System.out.println(t[0]+" "+t[1]);
	}
	
	public int[] compare(char[] a, char [] b){
		int[] ans= new int[2];
		int i=0;
		while(i!=a.length){
			if(a[i]!=b[i]){
				if(a[i]=='R'){
					if(b[i]=='S')ans[1]+=1;
					else ans[0]+=1;
				}
				else if(a[i]=='S'){
					if(b[i]=='P')ans[1]+=1;
					else ans[0]+=1;
			}
				else {
					if(b[i]=='R') ans[1]+=1;
					else ans[0]+=1;
				}
		}
			i++;
		}
		return ans;
		
	}
	
	public int gcd( int a, int b){
		if(a%b==0) return b;
		else return gcd(b,a%b);
		
	}
	
}

