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.*;
import java.io.*;
public class a {
public static void main(String[] args) throws IOException
{
    boolean[] sieve = new boolean[1000001];
    Arrays.fill(sieve, true);
    sieve[0]=sieve[1]=false;
    for(int i = 2; i<sieve.length; i++) if(sieve[i]) for(long j = (long)i*i; j<sieve.length; j+=i) sieve[(int)j] = false;
    //Scanner input = new Scanner(new File("input.txt"));
    //PrintWriter out = new PrintWriter(new File("output.txt"));
    PrintWriter out = new PrintWriter(System.out);
    input.init(System.in);
    int n = input.nextInt(), a =0, b=0;
    String s1 = input.next(), s2 = input.next();
    int product = s1.length()*s2.length();
    for(int i = 0; i<product; i++)
    {
        char c1 = s1.charAt(i%s1.length()), c2 = s2.charAt(i%s2.length());
        if(c1 == 'R' && c2 == 'P' || c1 == 'P' && c2 == 'S' || c1 == 'S' && c2 == 'R')a++;
        else if(c1 != c2) b++;
    }
    a *= n/product;
    b *= n/product;
    for(int i = 0; i<n%product; i++)
    {
        char c1 = s1.charAt(i%s1.length()), c2 = s2.charAt(i%s2.length());
        if(c1 == 'R' && c2 == 'P' || c1 == 'P' && c2 == 'S' || c1 == 'S' && c2 == 'R')a++;
        else if(c1 != c2) b++;
    }
    out.println(a+" "+b);
    out.close();
}

public static long gcd(long a, long b)
{
if(b == 0) return a;
return gcd(b, a%b);
}
public static class input {
    static BufferedReader reader;
    static StringTokenizer tokenizer;

    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
        reader = new BufferedReader(
                     new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }

    /** get next word */
    static String next() throws IOException {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                   reader.readLine() );
        }
        return tokenizer.nextToken();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt( next() );
    }
    
    static double nextDouble() throws IOException {
        return Double.parseDouble( next() );
    }
    static long nextLong() throws IOException {
        return Long.parseLong( next() );
    }
}
static class IT
{
    int[] left,right, val, a, b;
    IT(int n)
    {
        left = new int[3*n];
        right = new int[3*n];
        val = new int[3*n];
        a = new int[3*n];
        b = new int[3*n];
        init(0,0, n);
    }
    int init(int at, int l, int r)
    {
        a[at] = l;
        b[at] = r;
        if(l==r)
            left[at] = right [at] = -1;
        else 
        {
            int mid = (l+r)/2;
            left[at] = init(2*at+1,l,mid);
            right[at] = init(2*at+2,mid+1,r);
        }
        return at++;
    }
    //return the sum over [x,y]
    int get(int x, int y)
    {
        return go(x,y, 0);
    }
    int go(int x,int y, int at)
    {
        if(at==-1) return 0;
        if(x <= a[at] && y>= b[at]) return val[at];
        if(y<a[at] || x>b[at]) return 0;
        return go(x, y, left[at]) + go(x, y, right[at]);
    }
    //add v to elements x through y
    void add(int x, int y, int v)
    {
        go3(x, y, v, 0);
    }
    void go3(int x, int y, int v, int at)
    {
        if(at==-1) return;
        if(y < a[at] || x > b[at]) return;
        val[at] += (Math.min(b[at], y) - Math.max(a[at], x) + 1)*v;
        go3(x, y, v, left[at]);
        go3(x, y, v, right[at]);
    }
}
}
