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 main() {
  long long n;
  cin >> n;
  char a[1024], b[1024];
  cin >> a >> b;
  long a1 = strlen(a), b1 = strlen(b);
  long min = a1, nod = 1, max = b1;
  if (a1 > b1) {
    min = b1;
    max = a1;
  }
  for (long i = 2; i <= min; i++) {
    if ((max % i == 0) && (min % i == 0)) nod = i;
  }
  long nok = a1 * b1 / nod;
  long long p = n / nok;
  long long o = n % nok;
  long long nik = 0, pol1 = 0, nik1 = 0, pol = 0;
  if (n == 1) {
    if ((b[0] == 'R') && (a[0] == 'S')) {
      nik1++;
    }
    if ((b[0] == 'S') && (a[0] == 'P')) {
      nik1++;
    }
    if ((b[0] == 'P') && (a[0] == 'R')) {
      nik1++;
    }
    if ((a[0] == 'R') && (b[0] == 'S')) {
      pol1++;
    }
    if ((a[0] == 'S') && (b[0] == 'P')) {
      pol1++;
    }
    if ((a[0] == 'P') && (b[0] == 'R')) {
      pol1++;
    }
  } else {
    for (long long i = 0; i < nok; i++) {
      if ((a[i % a1] == 'R') && (b[i % b1] == 'S')) {
        pol++;
      }
      if ((a[i % a1] == 'S') && (b[i % b1] == 'P')) {
        pol++;
      }
      if ((a[i % a1] == 'P') && (b[i % b1] == 'R')) {
        pol++;
      }
      if ((b[i % b1] == 'R') && (a[i % a1] == 'S')) {
        nik++;
      }
      if ((b[i % b1] == 'S') && (a[i % a1] == 'P')) {
        nik++;
      }
      if ((b[i % b1] == 'P') && (a[i % a1] == 'R')) {
        nik++;
      }
    }
    nik1 = nik * p, pol1 = pol * p;
    pol = 0;
    nik = 0;
    for (long long i = n - o; i < n; i++) {
      if ((a[i % a1] == 'R') && (b[i % b1] == 'S')) {
        pol++;
      }
      if ((a[i % a1] == 'S') && (b[i % b1] == 'P')) {
        pol++;
      }
      if ((a[i % a1] == 'P') && (b[i % b1] == 'R')) {
        pol++;
      }
      if ((b[i % b1] == 'R') && (a[i % a1] == 'S')) {
        nik++;
      }
      if ((b[i % b1] == 'S') && (a[i % a1] == 'P')) {
        nik++;
      }
      if ((b[i % b1] == 'P') && (a[i % a1] == 'R')) {
        nik++;
      }
    }
    nik1 = nik1 + nik;
    pol1 = pol1 + pol;
  }
  cout << nik1 << " " << pol1;
  int z;
  cin >> z;
  return 0;
}
