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.Scanner;


public class A {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String s1 = sc.next();
        String s2 = sc.next();
        int a = s1.length(), b = s2.length(); 
        int g = gcd(a, b);
        int L = a*b / g;
        int aa = L / a, bb = L / b;
        char[] s11 = new char[L], s22 = new char[L];
        int k = 0;
        for (int i = 1; i <= aa; i++) {
            for (int j = 0; j < s1.length(); j++) {
                s11[k++] = s1.charAt(j);
            }
        }
        k = 0;
        for (int i = 1; i <= bb; i++) {
            for (int j = 0; j < s2.length(); j++) {
                s22[k++] = s2.charAt(j);
            }
        }
        int k1 = 0, k2 = 0;
        for (int i = 0; i < L; i++) {
            char w1 = s11[i], w2 = s22[i];
            if (w1=='R' && w2=='S' || w1=='S' && w2=='P' || w1=='P' && w2=='R')
                k1++;
            else {
                if (w1=='S' && w2=='R' || w1=='P' && w2=='S' || w1=='R' && w2=='P')
                    k2++;
            }
        }
        int m = n / L;
        int ans1 = k1 * m, ans2 = k2 * m;
        m = n % L;
        for (int i = 0; i < m; i++) {
            char w1 = s11[i], w2 = s22[i];
            if (w1=='R' && w2=='S' || w1=='S' && w2=='P' || w1=='P' && w2=='R')
                ans1++;
            else {
                if (w1=='S' && w2=='R' || w1=='P' && w2=='S' || w1=='R' && w2=='P')
                    ans2++;
            }
        }
        System.out.println(ans2+" "+ans1);
    }

    private static int gcd(int a, int b) {
        if (b==0)
            return a;
        return gcd(b, a % b);
    }
}
