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.Scanner;
import java.lang.Math;
public class Ishu
{
	public static void main(String[] args)
	{
	Scanner scan=new Scanner(System.in);
	long n,sa=0l,sb=0l,quo=1,rem=1;
	int hcf=1,i,j,k,lcm=1;
	String str1,str2;
	char[] A=new char[1000];
	char[] B=new char[1000];
	n=scan.nextLong();
	str1=scan.next();
	A=str1.toCharArray();
	str2=scan.next();
	B=str2.toCharArray();
	hcf=(str1.length()<=str2.length())?str1.length():str2.length();
	if(str1.length()%hcf==0&&str2.length()%hcf==0);
	else
		{
		for(hcf=(int)Math.sqrt((str1.length()<=str2.length())?(str1.length()):(str2.length()));hcf>1;--hcf)
			if(str1.length()%hcf==0&&str2.length()%hcf==0)
				break;
		}
	lcm=(str1.length()*str2.length())/hcf;
	quo=n/lcm;
	rem=n%lcm;
	j=k=0;
	for(i=1;i<=lcm;++i)
		{
		if(j==str1.length())
			j=0;
		if(k==str2.length())
			k=0;
		if(!(A[j]==B[k]))
			{
			if(A[j]=='R'&&B[k]=='S'||A[j]=='S'&&B[k]=='P'||A[j]=='P'&&B[k]=='R')
				++sb;
			else
				++sa;
			}
		++j;
		++k;
		}
	sa*=quo;
	sb*=quo;
	j=k=0;
	for(i=1;i<=rem;++i)
		{
		if(j==str1.length())
			j=0;
		if(k==str2.length())
			k=0;
		if(!(A[j]==B[k]))
			{
			if(A[j]=='R'&&B[k]=='S'||A[j]=='S'&&B[k]=='P'||A[j]=='P'&&B[k]=='R')
				++sb;
			else
				++sa;
			}
		++j;
		++k;
		}
	System.out.println(sa+" "+sb);
	}
}
