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;
int Gsd(int a, int b) { return (b > 0) ? Gsd(b, a % b) : a; }
int a, b;
string s1, s2;
int How_Much1(string s1, string s2, int Nok) {
  int answ1 = 0, j = 0, k = 0;
  for (int i = 0; i < Nok; i++) {
    if (s1[j] == 'P')
      if (s2[k] == 'S') answ1++;
    if (s1[j] == 'R')
      if (s2[k] == 'P') answ1++;
    if (s1[j] == 'S')
      if (s2[k] == 'R') answ1++;
    ++j;
    ++k;
    if (j >= a) j -= a;
    if (k >= b) k -= b;
  }
  return answ1;
}
int How_Much2(string s1, string s2, int Nok) {
  int answ2 = 0, j = 0, k = 0;
  for (int i = 0; i < Nok; i++) {
    if (s1[j] == 'P')
      if (s2[k] == 'R') answ2++;
    if (s1[j] == 'R')
      if (s2[k] == 'S') answ2++;
    if (s1[j] == 'S')
      if (s2[k] == 'P') answ2++;
    ++j;
    ++k;
    if (j >= a) j -= a;
    if (k >= b) k -= b;
  }
  return answ2;
}
int main() {
  int n;
  cin >> n;
  cin >> s1 >> s2;
  a = s1.size();
  b = s2.size();
  int NOK = a * b / Gsd(a, b);
  int answ1 = (n / NOK) * How_Much1(s1, s2, NOK), t = n % NOK, j = 0, k = 0;
  int answ2 = (n / NOK) * How_Much2(s1, s2, NOK);
  for (int i = 0; i < t; i++) {
    if (s1[j] == 'P') {
      if (s2[k] == 'R') answ2++;
      if (s2[k] == 'S') answ1++;
    }
    if (s1[j] == 'R') {
      if (s2[k] == 'S') answ2++;
      if (s2[k] == 'P') answ1++;
    }
    if (s1[j] == 'S') {
      if (s2[k] == 'R') answ1++;
      if (s2[k] == 'P') answ2++;
    }
    ++j;
    ++k;
    if (j >= a) j -= a;
    if (k >= b) k -= b;
  }
  cout << answ1 << ' ' << answ2;
  return 0;
}
