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.
#include <bits/stdc++.h>
int LCM(int a, int b);
int main() {
  int n;
  char a[10000], b[10000];
  int i, j, k;
  int x, y;
  int n1, n2;
  int s;
  int nb, ys;
  while (scanf("%d", &n) != EOF) {
    scanf("%s", &a);
    scanf("%s", &b);
    x = 0;
    y = 0;
    n1 = strlen(a);
    n2 = strlen(b);
    s = LCM(n1, n2);
    nb = n / s;
    ys = n % s;
    if (n >= s) {
      for (k = 0, i = 0, j = 0; k < s; i++, j++, k++) {
        i = i % n1;
        j = j % n2;
        if (a[i] == b[j]) continue;
        if (a[i] == 'R' && b[j] == 'P') x++;
        if (a[i] == 'P' && b[j] == 'S') x++;
        if (a[i] == 'S' && b[j] == 'R') x++;
        if (a[i] == 'R' && b[j] == 'S') y++;
        if (a[i] == 'S' && b[j] == 'P') y++;
        if (a[i] == 'P' && b[j] == 'R') y++;
      }
      x *= nb;
      y *= nb;
    }
    for (k = 0, i = 0, j = 0; k < ys; i++, j++, k++) {
      i = i % n1;
      j = j % n2;
      if (a[i] == b[j]) continue;
      if (a[i] == 'R' && b[j] == 'P') x++;
      if (a[i] == 'P' && b[j] == 'S') x++;
      if (a[i] == 'S' && b[j] == 'R') x++;
      if (a[i] == 'R' && b[j] == 'S') y++;
      if (a[i] == 'S' && b[j] == 'P') y++;
      if (a[i] == 'P' && b[j] == 'R') y++;
    }
    printf("%d %d\n", x, y);
  }
  return 0;
}
int LCM(int a, int b) {
  int c = a, d = b;
  int t;
  while (d != 0) {
    t = c % d;
    c = d;
    d = t;
  }
  return a / c * b;
}
