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.*;
import java.io.*;
import java.math.*;

public class Main {
	public static void main(String[] args) throws Exception {
		LR in = new LR();
		int n1 = Integer.parseInt(in.nx());
		char[] l1 = in.nx().toCharArray();
		char[] l2 = in.nx().toCharArray();
		long[][] q1 = new long[l1.length][l2.length];
		long[][] q2 = new long[l1.length][l2.length];
		long[][] d1 = new long[l1.length][l2.length];
		for (int i = 0;i<l1.length;i++){
			for (int j = 0;j<l2.length;j++){
				q1[i][j] = -1;
			}
		}
		for (int i = 0;i<l1.length;i++){
			for (int j = 0;j<l2.length;j++){
				q2[i][j] = -1;
			}
		}
		for (int i = 0;i<l1.length;i++){
			for (int j = 0;j<l2.length;j++){
				d1[i][j] = -1;
			}
		}
		int ptr1 = 0;
		int ptr2 = 0;
		long win1 = 0;
		long win2 = 0;
		for (long i =0;i<n1;i++){
			char[][] w1 = {{'R','S'},{'P','R'},{'S','P'}};
			for (char[] e: w1){
				if (l1[ptr1] == e[0] && l2[ptr2] == e[1]){
					win1++;
				}
			}
			for (char[] e: w1){
				if (l1[ptr1] == e[1] && l2[ptr2] == e[0]){
					win2++;
				}
			}
			
			if (q1[ptr1][ptr2] != -1){
				long last = d1[ptr1][ptr2];
				long len = i - last;
				long remain = n1 - 1 - i;
				if (remain / len >=1){
					long times = remain / len;
					i += times * len;
					long diff1 = win1 - q1[ptr1][ptr2];
					long diff2 = win2 - q2[ptr1][ptr2];
					win1 += times * diff1;
					win2 += times * diff2;
				}
				
			}
			q1[ptr1][ptr2] = win1;
			q2[ptr1][ptr2] = win2;
			d1[ptr1][ptr2] = i;
			ptr1 = (ptr1+1)%l1.length;
			ptr2 = (ptr2+1)%l2.length;
			//px(i,win1,win2);
		}
		
		pn(win2 +" "+win1);
	}

	private static BigInteger bi(long n1) {
		return BigInteger.valueOf(n1);
	}

	private static void p(Object o1) {
		System.out.print(o1);
	}

	private static void pn(Object o1) {
		System.out.println(o1);
	}

	private static void px(Object... o1) {
		System.out.println(Arrays.deepToString(o1));
	}

	private static class LR {
		BufferedReader k1 = null;

		public LR() throws Exception {
			k1 = new BufferedReader(new InputStreamReader(System.in));
		}

		public String nx() throws Exception {
			return k1.readLine();
		}
	}

	private static class ST {
		StreamTokenizer k1 = null;

		public ST() throws Exception {
			k1 = new StreamTokenizer(System.in);
		}

		public int ni() throws Exception {
			k1.nextToken();
			return (int) k1.nval;
		}

		public double nd() throws Exception {
			k1.nextToken();
			return (double) k1.nval;
		}

		public long nl() throws Exception {
			k1.nextToken();
			return (long) k1.nval;
		}
	}
}
