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 a, b, n, i, j, a2, b2;
char a1[1000009];
char b1[1000009];
int bet(char a3, char b3) {
  if ((a3 == 'R' && b3 == 'P') || (a3 == 'P' && b3 == 'S') ||
      (a3 == 'S' && b3 == 'R'))
    return a2++;
  if ((b3 == 'R' && a3 == 'P') || (b3 == 'P' && a3 == 'S') ||
      (b3 == 'S' && a3 == 'R'))
    return b2++;
  return 0;
}
int main() {
  scanf("%d\n%s\n%s", &n, a1, b1);
  a = strlen(a1);
  b = strlen(b1);
  for (i = a; i < b * a; i++) a1[i] = a1[i % a];
  for (i = b; i < a * b; i++) b1[i] = b1[i % b];
  for (i = 0; i < a * b; i++) bet(a1[i], b1[i]);
  a2 = a2 * (n / (a * b));
  b2 = b2 * (n / (a * b));
  for (i = 0; i < n % (a * b); i++) bet(a1[i], b1[i]);
  printf("%d %d\n", a2, b2);
}
