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 test
{

	public static void main(String[] args)
	{
		new test().run();
	}

	PrintWriter out = null;

	int isWin(char a, char b)
	{
		if (a == b)
			return 0;

		if (a == 'R')
		{
			if (b == 'S')
				return 1;
			else
				return -1;
		}

		if (a == 'S')
		{
			if (b == 'P')
				return 1;
			else
				return -1;
		}

		if (a == 'P')
		{
			if (b == 'R')
				return 1;
			else
				return -1;
		}

		return 0;
	}

	void run()
	{
		Scanner in = new Scanner(System.in);
		out = new PrintWriter(System.out);

		int n = in.nextInt();
		char[] p1 = in.next().toCharArray();
		char[] p2 = in.next().toCharArray();

		int l1 = p1.length;
		int l2 = p2.length;
		int m = l1 * l2;

		int[][] ret = new int[2][m + 1];

		for (int i = 1; i <= m; i++)
		{
			int rr = isWin(p1[(i - 1) % l1], p2[(i - 1) % l2]);
			if (rr == 0)
			{
				ret[0][i] = ret[0][i - 1];
				ret[1][i] = ret[1][i - 1];
			}
			else if (rr == 1)
			{
				ret[1][i] = 1 + ret[1][i - 1];
				ret[0][i] = ret[0][i - 1];
			}
			else
			{
				ret[0][i] = 1 + ret[0][i - 1];
				ret[1][i] = ret[1][i - 1];
			}
		}

		int[] counts = new int[2];

		int cc = n / m;
		counts[0] = ret[0][m] * cc;
		counts[1] = ret[1][m] * cc;

		if (n % m != 0)
		{
			cc = n % m;
			counts[0] += ret[0][cc];
			counts[1] += ret[1][cc];
		}

		out.println(counts[0] + " " + counts[1]);
		out.close();
	}
}