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, int);
int main() {
  int a[1000], b[1000], d = 0, x = 0, y = 0, z, k, i, j, e = 0, f = 0, n1, w;
  long int n;
  char s1[1001], s2[1001];
  scanf("%ld", &n);
  scanf("%s %s", s1, s2);
  int l1 = strlen(s1), l2 = strlen(s2);
  z = lcm(l1, l2);
  int m = (n) % z;
  n1 = (n - z) > 0 ? (n / z) : 1;
  w = (n - z) > 0 ? z : n;
  for (k = 1, j = 0, i = 0; k <= w; i++, j++, k++) {
    if (s1[i] == s2[j])
      d++;
    else if ((s1[i] == 'R' && s2[j] == 'S') || (s1[i] == 'S' && s2[j] == 'P') ||
             (s1[i] == 'P' && s2[j] == 'R'))
      x++;
    else
      y++;
    if ((k % l1) == 0) i = -1;
    if ((k % l2) == 0) j = -1;
    if (k == m && (n - z) > 0) {
      e = x;
      f = y;
    }
  }
  x = (x * n1) + e;
  y = (y * n1) + f;
  printf("%d %d ", y, x);
  return 0;
}
int lcm(int a, int b) {
  int i, h;
  h = a > b ? a : b;
  for (i = h; i <= (a * b); i += h) {
    if (i % a == 0 && i % b == 0) {
      return i;
    }
  }
}
