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.lang.*;
import java.math.BigInteger;
import java.io.*;
import java.util.*;

public class Solution implements Runnable{
    public static BufferedReader br;
    public static PrintWriter out;
    public static StringTokenizer stk;
    public static boolean isStream = true;

    public static void main(String[] args) throws IOException {
    	if (isStream) {
            br = new BufferedReader(new InputStreamReader(System.in));
        } else {
            br = new BufferedReader(new FileReader("in.txt"));
        }
        out = new PrintWriter(System.out);
        new Thread(new Solution()).start();
    }

    public void loadLine() {
        try {
            stk = new StringTokenizer(br.readLine());
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    public String nextLine() {
        try {
            return br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
            return "";
        }
    }

    public String nextWord() {
        while (stk==null||!stk.hasMoreTokens()) loadLine();
        return stk.nextToken();
    }

    public Integer nextInt() {
        while (stk==null||!stk.hasMoreTokens()) loadLine();
        return Integer.valueOf(stk.nextToken());
    }

    public Long nextLong() {
        while (stk==null||!stk.hasMoreTokens()) loadLine();
        return Long.valueOf(stk.nextToken());
    }

    public Double nextDouble() {
        while (stk==null||!stk.hasMoreTokens()) loadLine();
        return Double.valueOf(stk.nextToken());
    }
    
    public Float nextFloat() {
        while (stk==null||!stk.hasMoreTokens()) loadLine();
        return Float.valueOf(stk.nextToken());
    }
    
    int f(char a, char b) {
    	if (a == 'R' && b == 'S') return 0;
    	if (a == 'S' && b == 'R') return 1;
    	if (a == 'P' && b == 'R') return 0;
    	if (a == 'R' && b == 'P') return 1;
    	if (a == 'S' && b == 'P') return 0;
    	if (a == 'P' && b == 'S') return 1;
    	return -1;
    }
    
    public void run() {
    	int n = nextInt();
    	char[] a1, a2;
    	a1 = nextWord().toCharArray();
    	a2 = nextWord().toCharArray();
    	int m = a1.length;
    	int k = a2.length;
    	int[][] dp = new int[m][k];
    	for (int i = 0; i < m; i++) {
    		Arrays.fill(dp[i], -1);
    	}
    	int x = 0, y = 0;
    	int pos = 0;
    	int[] ans = new int[2];
    	while (dp[x][y] == -1 && pos < n) {
    		if (a1[x] != a2[y]) {
    			ans[f(a1[x], a2[y])]++;
    		}
    		dp[x][y] = pos;
    		x = (x+1)%m;
    		y = (y+1)%k;
    		pos++;
    	}
    	int cyc = pos - dp[x][y];
    	int div = (n-pos)/cyc;
    	int rem = (n-pos)%cyc;
    	for (int i = 0; i < cyc; i++) {
    		if (a1[x] != a2[y]) {
    			ans[f(a1[x], a2[y])] += div;
    		}
    		dp[x][y] = pos;
    		x = (x+1)%m;
    		y = (y+1)%k;
    		pos++;
    	}
    	for (int i = 0; i < rem; i++) {
    		if (a1[x] != a2[y]) {
    			ans[f(a1[x], a2[y])]++;
    		}
    		x = (x+1)%m;
    		y = (y+1)%k;
    		pos++;
    	}
    	out.println(ans[1] + " " + ans[0]);
    	out.flush();
    }
}
