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;
inline int read() {
  int ans = 0, flag = 1;
  char ch = getchar();
  while ((ch > '9' || ch < '0') && ch != '-') ch = getchar();
  if (ch == '-') flag = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') {
    ans = (ans << 1) + (ans << 3) + (ch ^ '0');
    ch = getchar();
  }
  return ans * flag;
}
int n, len, al, bl;
string a, b;
int gcd(int x, int y) { return !y ? x : gcd(y, x % y); }
inline int lcm(int x, int y) { return x * y / gcd(x, y); }
inline int findint(char u) { return (u == 'S' ? 1 : (u == 'P' ? 2 : 3)); }
inline int win(char x, char y) {
  int nowx = findint(x), nowy = findint(y);
  if (nowx == nowy) return 0;
  if (nowx == 1 && nowy == 2) return 1;
  if (nowx == 1 && nowy == 3) return 2;
  if (nowx == 2 && nowy == 3) return 1;
  if (nowx == 2 && nowy == 1) return 2;
  if (nowx == 3 && nowy == 2) return 2;
  if (nowx == 3 && nowy == 1) return 1;
}
int main() {
  n = read();
  cin >> a >> b;
  len = lcm(a.length(), b.length());
  for (int i = 0; i < len; i++)
    if (win(a[i % (a.length())], b[i % (b.length())]) == 1)
      bl++;
    else if (win(a[i % (a.length())], b[i % (b.length())]) == 2)
      al++;
  al *= (n / len);
  bl *= (n / len);
  for (int i = 0; i < n % len; i++)
    if (win(a[i % (a.length())], b[i % (b.length())]) == 1)
      bl++;
    else if (win(a[i % (a.length())], b[i % (b.length())]) == 2)
      al++;
  printf("%d %d\n", al, bl);
  return 0;
}
