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.*;
import static java.lang.Math.*;

public class CROC2012_Round1_A implements Runnable {

	final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
	
	BufferedReader in;
	PrintWriter out;
	StringTokenizer tok;

	public static void main(String[] args) {
		new Thread(null, new CROC2012_Round1_A(), "", 256 * (1L << 20)).start();
	}

	@Override
	public void run() {
		try {
			long startTime = System.currentTimeMillis();
			if (ONLINE_JUDGE) {
				in = new BufferedReader(new InputStreamReader(System.in));
				out = new PrintWriter(System.out);
			} else {
				in = new BufferedReader(new FileReader("input.txt"));
				out = new PrintWriter("output.txt");
			}
			Locale.setDefault(Locale.US);
			tok = new StringTokenizer("");
			solve();
			in.close();
			out.close();
			long freeMemory = Runtime.getRuntime().freeMemory();
			long totalMemory = Runtime.getRuntime().totalMemory();
			long endTime = System.currentTimeMillis();
			System.err.println("Time = " + (endTime - startTime));
			System.err.println("Memory = " + ((totalMemory - freeMemory) >> 10));
		} catch (Throwable t) {
			t.printStackTrace(System.err);
			System.exit(-1);
		}
	}

	String readString() throws IOException {
		while (!tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		return tok.nextToken();
	}

	int readInt() throws IOException {
		return Integer.parseInt(readString());
	}

	long readLong() throws IOException {
		return Long.parseLong(readString());
	}

	double readDouble() throws IOException {
		return Double.parseDouble(readString());
	}
	
	void debug(Object... o) {
		if (!ONLINE_JUDGE) {
			System.err.println(Arrays.deepToString(o));
		}
	}

	/** http://pastebin.com/j0xdUjDn */
	static class Utils {

		private Utils() {}

		public static void mergeSort(int[] a) {
			mergeSort(a, 0, a.length - 1);
		}

		private static final int MAGIC_VALUE = 50;

		private static void mergeSort(int[] a, int leftIndex, int rightIndex) {
			if (leftIndex < rightIndex) {
				if (rightIndex - leftIndex <= MAGIC_VALUE) {
					insertionSort(a, leftIndex, rightIndex);
				} else {
					int middleIndex = (leftIndex + rightIndex) / 2;
					mergeSort(a, leftIndex, middleIndex);
					mergeSort(a, middleIndex + 1, rightIndex);
					merge(a, leftIndex, middleIndex, rightIndex);
				}
			}
		}

		private static void merge(int[] a, int leftIndex, int middleIndex, int rightIndex) {
			int length1 = middleIndex - leftIndex + 1;
			int length2 = rightIndex - middleIndex;
			int[] leftArray = new int[length1];
			int[] rightArray = new int[length2];
			System.arraycopy(a, leftIndex, leftArray, 0, length1);
			System.arraycopy(a, middleIndex + 1, rightArray, 0, length2);
			for (int k = leftIndex, i = 0, j = 0; k <= rightIndex; k++) {
				if (i == length1) {
					a[k] = rightArray[j++];
				} else if (j == length2) {
					a[k] = leftArray[i++];
				} else {
					a[k] = leftArray[i] <= rightArray[j] ? leftArray[i++] : rightArray[j++];
				}
			}
		}

		private static void insertionSort(int[] a, int leftIndex, int rightIndex) {
			for (int i = leftIndex + 1; i <= rightIndex; i++) {
				int current = a[i];
				int j = i - 1;
				while (j >= leftIndex && a[j] > current) {
					a[j + 1] = a[j];
					j--;
				}
				a[j + 1] = current;
			}
		}

	}

	// solution
	
	void solve() throws IOException {
		int n = readInt();
		char[] a = readString().toCharArray();
		char[] b = readString().toCharArray();
		int T = a.length * b.length;
		int times = n / T;
		int stupid = n % T;
		long ans1 = 0, ans2 = 0;
		for (int i = 0; i < T; i++) {
			char ac = a[i % a.length];
			char bc = b[i % b.length];
			if (wins(ac, bc)) ans2++;
			if (wins(bc, ac)) ans1++;
		}
		ans1 *= times;
		ans2 *= times;
		for (int i = 0; i < stupid; i++) {
			char ac = a[i % a.length];
			char bc = b[i % b.length];
			if (wins(ac, bc)) ans2++;
			if (wins(bc, ac)) ans1++;
		}
		out.println(ans1 + " " + ans2);
	}

	private boolean wins(char win, char lose) {
		if (win == 'R' && lose == 'S') return true;
		if (win == 'S' && lose == 'P') return true;
		if (win == 'P' && lose == 'R') return true;
		return false;
	}

}