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;
long long min(long long a, long long b) {
  if (a > b) {
    return b;
  } else {
    return a;
  }
}
void bubbleSort(long long* array, long long length) {
  long long i, j;
  for (i = 0; i < length; i++) {
    for (j = 0; j < i; j++) {
      if (array[i] < array[j]) {
        long long temp = array[i];
        array[i] = array[j];
        array[j] = temp;
      }
    }
  }
}
void change(char* aa, char* bb, int* a, int* b) {
  if (*aa != *bb) {
    if (((*aa == 'R') && (*bb == 'P')) || ((*aa == 'P') && (*bb == 'S')) ||
        ((*aa == 'S') && (*bb == 'R'))) {
      (*b)++;
    } else {
      (*a)++;
    }
  }
}
int main() {
  int n;
  cin >> n;
  string a1, b1;
  cin >> a1 >> b1;
  int a = 0;
  int b = 0;
  int i = 0;
  int j = 0;
  for (int k = 0; k < min(a1.size() * b1.size(), n); k++) {
    change(&a1[i], &b1[j], &a, &b);
    i = (i + 1) % a1.size();
    j = (j + 1) % b1.size();
  }
  a *= n / (min(a1.size() * b1.size(), n));
  b *= n / (min(a1.size() * b1.size(), n));
  i = 0;
  j = 0;
  for (int k = (n / (min(a1.size() * b1.size(), n))) * a1.size() * b1.size();
       k < n; k++) {
    change(&(a1[i]), &(b1[j]), &a, &b);
    i = (i + 1) % a1.size();
    j = (j + 1) % b1.size();
  }
  cout << b << " " << a;
}
