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 mcd(int a, int b) {
  int c;
  while (a != b) {
    if (a > b)
      a = a - b;
    else
      b = b - a;
  }
  c = a;
  return c;
}
void comp(char A, char B, long long cont1, long long cont2) {
  if (A == 'P' && B == 'S')
    cont1++;
  else if (A == 'S' && B == 'P')
    cont2++;
  else if (A == 'R' && B == 'P')
    cont1++;
  else if (A == 'P' && B == 'R')
    cont2++;
  else if (A == 'S' && B == 'R')
    cont1++;
  else if (A == 'R' && B == 'S')
    cont2++;
}
int main() {
  long long n;
  cin >> n;
  string A, B;
  cin >> A >> B;
  string aux = A, aux2 = B;
  int num = (A.size() * B.size()) / mcd(A.size(), B.size());
  for (int i = 1; i < num / aux.size(); i++) A += aux;
  A += aux.substr(0, num % aux.size());
  for (int i = 1; i < num / aux2.size(); i++) B += aux2;
  B += aux2.substr(0, num % aux2.size());
  long long cont1 = 0, cont2 = 0;
  if (n <= num) {
    for (int i = 0; i < n; i++) {
      if (A[i] == 'P' && B[i] == 'S')
        cont1++;
      else if (A[i] == 'S' && B[i] == 'P')
        cont2++;
      else if (A[i] == 'R' && B[i] == 'P')
        cont1++;
      else if (A[i] == 'P' && B[i] == 'R')
        cont2++;
      else if (A[i] == 'S' && B[i] == 'R')
        cont1++;
      else if (A[i] == 'R' && B[i] == 'S')
        cont2++;
    }
  } else {
    for (int i = 0; i < num; i++) {
      if (A[i] == 'P' && B[i] == 'S')
        cont1++;
      else if (A[i] == 'S' && B[i] == 'P')
        cont2++;
      else if (A[i] == 'R' && B[i] == 'P')
        cont1++;
      else if (A[i] == 'P' && B[i] == 'R')
        cont2++;
      else if (A[i] == 'S' && B[i] == 'R')
        cont1++;
      else if (A[i] == 'R' && B[i] == 'S')
        cont2++;
    }
    cont1 = cont1 * (n / num);
    cont2 = cont2 * (n / num);
    for (int i = 0; i < n % num; i++) {
      if (A[i] == 'P' && B[i] == 'S')
        cont1++;
      else if (A[i] == 'S' && B[i] == 'P')
        cont2++;
      else if (A[i] == 'R' && B[i] == 'P')
        cont1++;
      else if (A[i] == 'P' && B[i] == 'R')
        cont2++;
      else if (A[i] == 'S' && B[i] == 'R')
        cont1++;
      else if (A[i] == 'R' && B[i] == 'S')
        cont2++;
    }
  }
  cout << cont1 << " " << cont2 << endl;
  return 0;
}
