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>
using namespace std;
void read(int &k) {
  k = 0;
  char x = getchar();
  while (x < '0' || x > '9') x = getchar();
  while (x >= '0' && x <= '9') {
    k = k * 10 - 48 + x;
    x = getchar();
  }
}
char s1[1000010], s2[1000010];
int N, a[1000010], b[1000010], n, m;
bool win(char a, char b) {
  return (a == 'R' && b == 'S') || (a == 'S' && b == 'P') ||
         (a == 'P' && b == 'R');
}
int main() {
  scanf("%d%s%s", &N, s1, s2);
  n = strlen(s1);
  m = strlen(s2);
  for (int i = 1; i < m; i++)
    for (int j = 0; j < n; j++) s1[i * n + j] = s1[j];
  for (int i = 1; i < n; i++)
    for (int j = 0; j < m; j++) s2[i * m + j] = s2[j];
  a[0] = b[0] = 0;
  for (int i = 0; i < n * m; i++) {
    if (i) a[i] = a[i - 1], b[i] = b[i - 1];
    if (win(s1[i], s2[i])) a[i]++;
    if (win(s2[i], s1[i])) b[i]++;
  }
  int p = N / (n * m), q = N % (n * m);
  int ans1 = p * a[n * m - 1];
  if (q) ans1 += a[q - 1];
  int ans2 = p * b[n * m - 1];
  if (q) ans2 += b[q - 1];
  printf("%d %d\n", ans2, ans1);
  return 0;
}
