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;

/**
 * TODO: class description.
 * Date: 06.03.12
 * Time: 18:30
 *
 * @author German Sysoev (HSE)
 */
public class A {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        char[] p1 = sc.next().toCharArray();
        char[] p2 = sc.next().toCharArray();
        int[] result = new int[p1.length * p2.length];
        int c1s = 0;
        int c2s = 0;

        for (int i = 0, k = result.length, i1 = -1, i2 = -1; i < k; ++i) {
            if (i1 + 1 == p1.length) {
                i1 = 0;
            } else {
                i1++;
            }
            if (i2 + 1 == p2.length) {
                i2 = 0;
            } else {
                i2++;
            }
            if (p1[i1] == p2[i2]) {
                result[i] = 0;
            } else if (p1[i1] == 'P' && p2[i2] == 'R' ||
                    p1[i1] == 'R' && p2[i2] == 'S' ||
                    p1[i1] == 'S' && p2[i2] == 'P'){
                result[i] = -1;
                c1s++;
            } else {
                result[i] = 1;
                c2s++;
            }
        }

        int r1 = (n / result.length) * c1s;
        int r2 = (n / result.length) * c2s;
        int tmp = n - (n / result.length) * result.length;
        for (int i = 0; i < tmp; i++) {
            if (result[i] == -1)
                r1++;
            else if (result[i] == 1)
                r2++;
        }

        System.out.print(r2 + " " + r1);
    }
}
