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 int LCM(int, int);
bool bigger(const char &, const char &, bool &);
int main() {
  long int n, _lcm;
  int lena, lenb;
  bool flag;
  string a, b;
  cin >> n >> a >> b;
  lena = a.length();
  lenb = b.length();
  _lcm = LCM(lena, lenb);
  long int suma = 0, sumb = 0, cnt = 0;
  int i(0), j(0);
  while (cnt < n && cnt < _lcm) {
    flag = false;
    if (bigger(a[i++], b[j++], flag))
      suma++;
    else if (!flag)
      sumb++;
    cnt++;
    if (i == lena) i = 0;
    if (j == lenb) j = 0;
  }
  long int t = n / _lcm;
  if (t == 0)
    cout << sumb << ' ' << suma << endl;
  else {
    suma *= t;
    sumb *= t;
    cnt = t * _lcm;
    i = j = 0;
    while (cnt < n) {
      flag = false;
      if (bigger(a[i++], b[j++], flag))
        suma++;
      else if (!flag)
        sumb++;
      cnt++;
      if (i == lena) i = 0;
      if (j == lenb) j = 0;
    }
    cout << sumb << ' ' << suma << endl;
  }
  return 0;
}
long int LCM(int a, int b) {
  if (a < b) {
    int temp = a;
    a = b;
    b = temp;
  }
  int k = a;
  while (a % b != 0) a += k;
  return a;
}
bool bigger(const char &a, const char &b, bool &f) {
  if (a == b) {
    f = true;
    return false;
  }
  switch (a) {
    case 'R':
      if (b == 'S') return true;
      if (b == 'P') return false;
      break;
    case 'S':
      if (b == 'R') return false;
      if (b == 'P') return true;
      break;
    case 'P':
      if (b == 'R') return true;
      if (b == 'S') return false;
  }
}
