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 Solution {

    BufferedReader in;
    StringTokenizer st;
    PrintWriter out;

    int n;
    String s1, s2;

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

    int lca(int a, int b) {
        return a * b / gcd(a, b);
    }

    int count(String s1, String s2, int last) {
        int ret = 0;
        for (int i = 0; i < last; ++i) {
            char c1 = s1.charAt(i);
            char c2 = s2.charAt(i);
            if (check(c1, c2))
                ++ret;
        }
        return ret;
    }

    boolean check(char c1, char c2) {
        if (c1 == 'R' && c2 == 'S') {
            return true;
        }
        if (c1 == 'S' && c2 == 'P')
            return true;
        if (c1 == 'P' && c2 == 'R')
            return true;
        return false;
    }

    void solve() throws IOException {
        n = ni();
        s1 = ns();
        s2 = ns();
        int lca = lca(s1.length(), s2.length());
        int need1 = lca / s1.length();
        int need2 = lca / s2.length();
        s1 = inc(s1, need1);
        s2 = inc(s2, need2);

        int times = n / lca;
        int mod = n % lca;
        int a = count(s1, s2, s1.length()) * times + count(s1, s2, mod);
        int b = count(s2, s1, s2.length()) * times + count(s2, s1, mod);
        out.println(b + " " + a);
    }

    String inc(String s, int cnt) {
        StringBuilder ret = new StringBuilder(s);
        for (int i = 2; i <= cnt; ++i)
            ret.append(s);
        return ret.toString();
    }

    public Solution() throws IOException {
        // int max = 0;
        // int i1 = 0, j1 = 0;
        // for (int i = 1; i <= 1000; ++i)
        // for (int j = 1; j <= 1000; ++j) {
        // if (lca(i, j) > max) {
        // max = lca(i, j);
        // i1 = i;
        // j1 = j;
        // }
        // max = max(max, lca(i, j));
        // }
        // System.err.println(max + " " + i1 + " " + j1);
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        in.close();
        out.close();
    }

    String ns() throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(in.readLine());
        return st.nextToken();
    }

    int ni() throws IOException {
        return Integer.valueOf(ns());
    }

    long nl() throws IOException {
        return Long.valueOf(ns());
    }

    double nd() throws IOException {
        return Double.valueOf(ns());
    }

    public static void main(String[] args) throws IOException {
        new Solution();
    }
}
