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;
template <typename T>
inline T read() {
  T f = 1, x = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
string s1, s2;
int n, loses[1000007], loset[1000007], lose1, lose2, ans1, ans2;
inline int gcd(int a, int b) { return !b ? a : gcd(b, a % b); }
inline int lcm(int a, int b) { return a / gcd(a, b) * b; }
int main() {
  n = read<int>();
  cin >> (s1);
  int(len1) = (s1).size();
  cin >> (s2);
  int(len2) = (s2).size();
  int round = lcm(len1, len2);
  for (int(i) = (0); (i) <= (round - 1); ++(i)) {
    if ((s1[i % len1] == 'R' && s2[i % len2] == 'P') ||
        (s1[i % len1] == 'P' && s2[i % len2] == 'S') ||
        (s1[i % len1] == 'S' && s2[i % len2] == 'R'))
      loses[i + 1]++, lose1++;
    if ((s2[i % len2] == 'R' && s1[i % len1] == 'P') ||
        (s2[i % len2] == 'P' && s1[i % len1] == 'S') ||
        (s2[i % len2] == 'S' && s1[i % len1] == 'R'))
      loset[i + 1]++, lose2++;
  }
  for (int(i) = (1); (i) <= (n % round); ++(i))
    ans1 += loses[i], ans2 += loset[i];
  return printf("%d %d", ans1 + lose1 * (n / round),
                ans2 + lose2 * (n / round)),
         0;
}
