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;
double pi = 3.141592653589793;
int main() {
  long long n, i;
  string str1, str2;
  cin >> n;
  cin >> str1 >> str2;
  long long l1 = str1.length(), l2 = str2.length(), one = 0, two = 0,
            count1 = 0, count2 = 0;
  for (i = 0; i < min(l1 * l2, n); ++i) {
    if (str1[one] == 'R' && str2[two] == 'S')
      count1++;
    else if (str1[one] == 'S' && str2[two] == 'R')
      count2++;
    else if (str1[one] == 'P' && str2[two] == 'R')
      count1++;
    else if (str1[one] == 'R' && str2[two] == 'P')
      count2++;
    else if (str1[one] == 'P' && str2[two] == 'S')
      count2++;
    else if (str1[one] == 'S' && str2[two] == 'P')
      count1++;
    if (one == l1 - 1)
      one = 0;
    else
      one++;
    if (two == l2 - 1)
      two = 0;
    else
      two++;
  }
  if (l1 * l2 < n) {
    n -= l1 * l2;
    int x = (n / (l1 * l2));
    if (x >= 1) {
      count1 = count1 + count1 * x;
      count2 = count2 + count2 * x;
    }
    n = n % (l1 * l2);
    for (i = 0; i < n; ++i) {
      if (str1[one] == 'R' && str2[two] == 'S')
        count1++;
      else if (str1[one] == 'S' && str2[two] == 'R')
        count2++;
      else if (str1[one] == 'P' && str2[two] == 'R')
        count1++;
      else if (str1[one] == 'R' && str2[two] == 'P')
        count2++;
      else if (str1[one] == 'P' && str2[two] == 'S')
        count2++;
      else if (str1[one] == 'S' && str2[two] == 'P')
        count1++;
      if (one == l1 - 1)
        one = 0;
      else
        one++;
      if (two == l2 - 1)
        two = 0;
      else
        two++;
    }
  }
  cout << count2 << " " << count1;
  return 0;
}
