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 main() {
  int n, a = 0, b = 0, i, l2, l1, flag = 1, s1 = 0, s2 = 0, T;
  char str1[1010], str2[1010];
  scanf("%d", &n);
  scanf("%s%s", str1, str2);
  l1 = strlen(str1);
  l2 = strlen(str2);
  for (i = 1; i <= n; i++) {
    if (a == 0 && b == 0 && i != 1 && flag == 1) {
      flag = 0;
      T = n / (i - 1);
      s1 = s1 * T;
      s2 = s2 * T;
      i = (i - 1) * T;
      continue;
    }
    if (str1[a] == str2[b])
      ;
    else if (str1[a] == 'S' && str2[b] == 'P')
      s1++;
    else if (str1[a] == 'S' && str2[b] == 'R')
      s2++;
    else if (str1[a] == 'R' && str2[b] == 'P')
      s2++;
    else if (str1[a] == 'R' && str2[b] == 'S')
      s1++;
    else if (str1[a] == 'P' && str2[b] == 'S')
      s2++;
    else if (str1[a] == 'P' && str2[b] == 'R')
      s1++;
    a++;
    b++;
    if (a == l1) a = a % l1;
    if (b == l2) b = b % l2;
  }
  printf("%d %d\n", s2, s1);
  return 0;
}
