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 fun(int a, int b) {
  int tmp;
  while (a % b != 0) {
    tmp = a % b;
    a = b;
    b = tmp;
  }
  return b;
}
int main() {
  long long int n;
  string str1, str2;
  long long int count1, count2;
  long long int lcm, tmp, mn;
  count1 = 0;
  count2 = 0;
  char ch2, ch1;
  cin >> n >> str1 >> str2;
  long long int si1 = str1.size();
  long long int si2 = str2.size();
  lcm = (si1 * si2) / fun(si1, si2);
  tmp = n % lcm;
  mn = min(n, lcm);
  for (int i = 0; i < mn; i++) {
    if ((i % si1) >= 0 && (i % si1) < (str1.size())) {
      ch1 = str1[i % si1];
    }
    if ((i % si2) >= 0 && (i % si2) < (str2.size())) {
      ch2 = str2[i % si2];
    }
    if (ch1 == 'R' && ch2 == 'S')
      count1++;
    else if (ch1 == 'S' && ch2 == 'R')
      count2++;
    else if (ch1 == 'P' && ch2 == 'S')
      count2++;
    else if (ch1 == 'S' && ch2 == 'P')
      count1++;
    else if (ch1 == 'P' && ch2 == 'R')
      count1++;
    else if (ch1 == 'R' && ch2 == 'P')
      count2++;
  }
  if (mn == lcm) {
    count1 = count1 * (n / lcm);
    count2 = count2 * (n / lcm);
    for (int i = 0; i < tmp; i++) {
      if ((i % si1) >= 0 && (i % si1) < (str1.size())) {
        ch1 = str1[i % si1];
      }
      if ((i % si2) >= 0 && (i % si2) < (str2.size())) {
        ch2 = str2[i % si2];
      }
      if (ch1 == 'R' && ch2 == 'S')
        count1++;
      else if (ch1 == 'S' && ch2 == 'R')
        count2++;
      else if (ch1 == 'P' && ch2 == 'S')
        count2++;
      else if (ch1 == 'S' && ch2 == 'P')
        count1++;
      else if (ch1 == 'P' && ch2 == 'R')
        count1++;
      else if (ch1 == 'R' && ch2 == 'P')
        count2++;
    }
  }
  cout << count2 << " " << count1 << endl;
  return 0;
}
