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.*;
public class A {
	static char[] a = new char[2001], b  = new char[2001];
	static int n,m,k;
	static long ans1=0,ans2=0;
	static void f(int i,int j){
		if (a[i]=='R' && b[j]=='S') ans1++; else
			if (a[i]=='P' && b[j]=='R') ans1++; else
				if (a[i]=='S' && b[j]=='P') ans1++; else
					
					if (b[j]=='R' && a[i]=='S') ans2++; else
						if (b[j]=='P' && a[i]=='R') ans2++; else
							if (b[j]=='S' && a[i]=='P') ans2++; 			
	}
	
public static void main(String args[]) {
	Scanner in = new Scanner(System.in);
	n=in.nextInt(); 
 String yed=in.next(); m =yed.length();for (int j=1;j<=m;j++) a[j] = yed.charAt(j-1);
 yed = in.next(); k =yed.length();for (int j=1;j<=k;j++) b[j] = yed.charAt(j-1);
boolean[][] use  = new boolean[2001][2001];
 int i=1,j=1,len=0;;
 while (!use[i][j] && len<n) {
	 use[i][j]=true;f(i,j);
	 len++;
	 
	 i++;if (i>m) i = i-m;
	 j++;if (j>k) j = j-k; 
	 
 }
 
 ans1 = ans1*(n/len);
 ans2 = ans2*(n/len);
 
 
 
 for (int t=1;t<=n%len;t++) {
	 f(i,j);
	 i++;if (i>m) i = i-m;
	 j++;if (j>k) j = j-k;  
 }
 
System.out.println(ans2+" "+ans1);

 }
}
