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 long int lcm(long long int i, long long int j) {
  int k;
  if (i == j)
    return i;
  else if (i > j) {
    for (k = i; k <= i * j; k++) {
      if (k % i == 0 && k % j == 0) return k;
    }
  } else {
    for (k = j; k <= i * j; k++) {
      if (k % i == 0 && k % j == 0) return k;
    }
  }
}
int main() {
  long long int n, i, j, k, no, rem, ls1, ls2, boy1 = 0, boy2 = 0, boy3, boy4,
                                               lc;
  char x1, x2, s1[1010], s2[1010];
  cin >> n;
  scanf("%s", s1);
  scanf("%s", s2);
  ls1 = strlen(s1);
  ls2 = strlen(s2);
  {
    long long int min;
    lc = lcm(ls1, ls2);
    no = n / lc;
    rem = n % lc;
    if (lc > n)
      min = n;
    else
      min = lc;
    j = 0;
    k = 0;
    if (no != 0) {
      boy1 = 0;
      boy2 = 0;
      for (i = 0; i < min; i++) {
        x1 = s1[j];
        x2 = s2[k];
        if (x1 == 'R' && x2 == 'S') boy2++;
        if (x1 == 'S' && x2 == 'R') boy1++;
        if (x1 == 'S' && x2 == 'P') boy2++;
        if (x2 == 'S' && x1 == 'P') boy1++;
        if (x1 == 'P' && x2 == 'R') boy2++;
        if (x2 == 'P' && x1 == 'R') boy1++;
        j++;
        k++;
        if (j == ls1) j = 0;
        if (k == ls2) k = 0;
      }
    }
    j = 0;
    k = 0;
    boy3 = 0;
    boy4 = 0;
    for (i = 0; i < rem; i++) {
      x1 = s1[j];
      x2 = s2[k];
      if (x1 == 'R' && x2 == 'S')
        boy4 = boy4 + 1;
      else if (x1 == 'S' && x2 == 'R')
        boy3 = boy3 + 1;
      else if (x1 == 'S' && x2 == 'P')
        boy4 = boy4 + 1;
      else if (x2 == 'S' && x1 == 'P')
        boy3 = boy3 + 1;
      else if (x1 == 'P' && x2 == 'R')
        boy4 = boy4 + 1;
      else if (x2 == 'P' && x1 == 'R')
        boy3 = boy3 + 1;
      j++;
      k++;
      if (j == ls1) j = 0;
      if (k == ls2) k = 0;
    }
    cout << (no * boy1 + boy3) << " " << (no * boy2 + boy4);
  }
  return 0;
}
