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 i, j, k, n, m;
  int ind, ind1, ind2, r1 = 0, r2 = 0;
  ind = 0;
  ind1 = 0;
  ind2 = 0;
  string a, b, c, d;
  cin >> n;
  cin >> c;
  cin >> d;
  int f;
  int l1 = c.length();
  int l2 = d.length();
  if (l1 >= l2) {
    f = 0;
    a = c;
    b = d;
  } else {
    int temp = l2;
    l2 = l1;
    l1 = temp;
    f = 1;
    a = d;
    b = c;
  }
  while (ind < n) {
    if (a[ind1] != b[ind2]) {
      if (a[ind1] == 'R') {
        if (b[ind2] == 'P')
          r1++;
        else
          r2++;
      } else if (a[ind1] == 'S') {
        if (b[ind2] == 'R')
          r1++;
        else
          r2++;
      } else {
        if (b[ind2] == 'S')
          r1++;
        else
          r2++;
      }
    }
    ind1++;
    ind2++;
    ind1 = ind1 % l1;
    ind2 = ind2 % l2;
    ind++;
    if (ind1 == 0 && ind2 == 0) {
      break;
    }
  }
  int num = n / ind;
  int num1 = n % ind;
  if (num > 0) {
    r1 = r1 * num;
    r2 = r2 * num;
    ind = 0;
    ind1 = 0;
    ind2 = 0;
    while (1 && ind < num1) {
      if (a[ind1] != b[ind2]) {
        if (a[ind1] == 'R') {
          if (b[ind2] == 'P')
            r1++;
          else
            r2++;
        } else if (a[ind1] == 'S') {
          if (b[ind2] == 'R')
            r1++;
          else
            r2++;
        } else {
          if (b[ind2] == 'S')
            r1++;
          else
            r2++;
        }
      }
      ind1++;
      ind2++;
      ind1 = ind1 % l1;
      ind2 = ind2 % l2;
      ind++;
      if (ind1 == 0 && ind2 == 0) {
        break;
      }
    }
  }
  if (f == 0)
    cout << r1 << " " << r2 << "\n";
  else
    cout << r2 << " " << r1 << "\n";
  return 0;
}
