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 int n;
  cin >> n;
  double hi = n;
  string w1;
  string w2;
  cin >> w1;
  cin >> w2;
  long long int gk = w1.length() * w2.length();
  string w3 = w1;
  string w4 = w2;
  long long int jkkl = w1.length() * w2.length();
  double bn = w1.length();
  double bn2 = w2.length();
  double ls = jkkl / bn;
  double ls2 = jkkl / bn2;
  for (long long int i = 1; i < ls; i++) {
    w1 = w1 + w3;
  }
  for (long long int i = 1; i < ls2; i++) {
    w2 = w2 + w4;
  }
  string w10 = w1.substr(0, jkkl);
  string w11 = w2.substr(0, jkkl);
  long long int pinch1 = 0;
  long long int pinch2 = 0;
  for (long long int i = 0; i < jkkl; i++) {
    if (w10[i] == w11[i]) {
      continue;
    }
    if (w10[i] == 'R' && w11[i] == 'S') {
      pinch2++;
    } else if (w10[i] == 'S' && w11[i] == 'R') {
      pinch1++;
    } else if (w10[i] == 'S' && w11[i] == 'P') {
      pinch2++;
    } else if (w10[i] == 'P' && w11[i] == 'S') {
      pinch1++;
    } else if (w10[i] == 'R' && w11[i] == 'P') {
      pinch1++;
    } else if (w10[i] == 'P' && w11[i] == 'R') {
      pinch2++;
    }
  }
  long long int hame = n / jkkl;
  pinch1 = pinch1 * hame;
  pinch2 = pinch2 * hame;
  if (hame * jkkl == n) {
    cout << pinch1 << " " << pinch2 << endl;
  } else {
    long long int mm = hame * jkkl;
    long long int dif = abs(mm - n);
    for (long long int i = 0; i < dif; i++) {
      if (w10[i] == w11[i]) {
        continue;
      }
      if (w10[i] == 'R' && w11[i] == 'S') {
        pinch2++;
      } else if (w10[i] == 'S' && w11[i] == 'R') {
        pinch1++;
      } else if (w10[i] == 'S' && w11[i] == 'P') {
        pinch2++;
      } else if (w10[i] == 'P' && w11[i] == 'S') {
        pinch1++;
      } else if (w10[i] == 'R' && w11[i] == 'P') {
        pinch1++;
      } else if (w10[i] == 'P' && w11[i] == 'R') {
        pinch2++;
      }
    }
    cout << pinch1 << " " << pinch2 << endl;
  }
}
