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
{
	public static void main(String args[])
	{
		Scanner scan = new Scanner(System.in);

		int n = scan.nextInt();
		String str1 = scan.next();
		String str2 = scan.next();

		int cyc = str1.length() * str2.length();

		int w1 = 0;
		int w2 = 0;

		for(int i=0;i < cyc;i++)
		{
			char c1 = str1.charAt(i%str1.length());
			char c2 = str2.charAt(i%str2.length());

			if(c1 == c2) continue;
			if(c1 == 'R' && c2 == 'S') w1++;
			if(c1 == 'S' && c2 == 'R') w2++;
			if(c1 == 'S' && c2 == 'P') w1++;
			if(c1 == 'P' && c2 == 'S') w2++;
			if(c1 == 'P' && c2 == 'R') w1++;
			if(c1 == 'R' && c2 == 'P') w2++;
		}

		int rtn1 = 0;
		int rtn2 = 0;
		int k = n / cyc;
		rtn1 += w2 * k;
		rtn2 += w1 * k;

		n = n - (cyc*k);
		w1 = 0;
		w2 = 0;

		for(int i=0;i < n;i++)
		{
			char c1 = str1.charAt(i%str1.length());
			char c2 = str2.charAt(i%str2.length());

			if(c1 == c2) continue;
			if(c1 == 'R' && c2 == 'S') w1++;
			if(c1 == 'S' && c2 == 'R') w2++;
			if(c1 == 'S' && c2 == 'P') w1++;
			if(c1 == 'P' && c2 == 'S') w2++;
			if(c1 == 'P' && c2 == 'R') w1++;
			if(c1 == 'R' && c2 == 'P') w2++;
		}

		rtn1 += w2;
		rtn2 += w1;

		System.out.println(rtn1 + " " + rtn2);
	}
}