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 gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main(int argc, char* argv[]) {
  int n, schel_nik = 0, schel_pol = 0, n1, n2;
  scanf("%d", &n);
  char a[1010], b[1010];
  scanf("%s", a);
  scanf("%s", b);
  n1 = strlen(a);
  n2 = strlen(b);
  int i = 0, l, k;
  int q = 0, w = 0;
  int nok = (n1 * n2) / gcd(n1, n2);
  int ost = 0;
  if (nok < n) {
    l = nok;
    k = n / nok;
    ost = n - k * nok;
  } else {
    l = n;
    k = 1;
  }
  while (i != l) {
    if (a[q] == b[w]) {
      ++q;
      ++w;
    } else {
      switch (a[q]) {
        case 'R':
          if (b[w] == 'P')
            schel_pol++;
          else
            schel_nik++;
          break;
        case 'P':
          if (b[w] == 'S')
            schel_pol++;
          else
            schel_nik++;
          break;
        case 'S':
          if (b[w] == 'R')
            schel_pol++;
          else
            schel_nik++;
          break;
        default:;
      }
      ++q;
      ++w;
    }
    if (q == n1) q = 0;
    if (w == n2) w = 0;
    ++i;
  }
  schel_pol *= k;
  schel_nik *= k;
  i = 0;
  q = 0;
  w = 0;
  while (i != ost) {
    if (a[q] == b[w]) {
      ++q;
      ++w;
    } else {
      switch (a[q]) {
        case 'R':
          if (b[w] == 'P')
            schel_pol++;
          else
            schel_nik++;
          break;
        case 'P':
          if (b[w] == 'S')
            schel_pol++;
          else
            schel_nik++;
          break;
        case 'S':
          if (b[w] == 'R')
            schel_pol++;
          else
            schel_nik++;
          break;
        default:;
      }
      ++q;
      ++w;
    }
    if (q == n1) q = 0;
    if (w == n2) w = 0;
    ++i;
  }
  printf("%d %d", schel_pol, schel_nik);
  scanf("%d", &i);
  return 0;
}
