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.*;

/**
 * @author OleGG
 */
public class Task173A {

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

    public static void main(String[] args) throws Throwable{
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        char[] a = sc.next().toCharArray();
        char[] b = sc.next().toCharArray();

        int la = a.length;
        int lb = b.length;
        int cycleLen = la * lb / gcd(la, lb);

        int[][] r1 = new int[256][256];
        int[][] r2 = new int[256][256];
        for (int i = 0; i < 256; ++i) {
            Arrays.fill(r1[i], 0);
            Arrays.fill(r2[i], 0);
        }
        r1['R']['S'] = 1;
        r1['S']['P'] = 1;
        r1['P']['R'] = 1;

        r2['S']['R'] = 1;
        r2['P']['S'] = 1;
        r2['R']['P'] = 1;

        int res1 = 0, res2 = 0;

        for (int i = 0; i < cycleLen; ++i) {
            res1 += r1[a[i % la]][b[i % lb]];
            res2 += r2[a[i % la]][b[i % lb]];
        }

        res1 *= n / cycleLen;
        res2 *= n / cycleLen;

        for (int i = 0; i < n % cycleLen; ++i) {
            res1 += r1[a[i % la]][b[i % lb]];
            res2 += r2[a[i % la]][b[i % lb]];
        }

        System.out.println(res2 + " " + res1);
    }
}
