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>
char a[1003], b[1003];
int gcd(int a, int b) {
  while (a && b) {
    if (a > b)
      a %= b;
    else
      b %= a;
  }
  return a + b;
}
inline int cmp(char a, char b) {
  if (a == 'R') {
    if (b == 'S')
      return 1;
    else if (b == 'P')
      return -1;
    else
      return 0;
  } else if (a == 'S') {
    if (b == 'R')
      return -1;
    else if (b == 'P')
      return 1;
    else
      return 0;
  } else {
    if (b == 'S')
      return -1;
    else if (b == 'R')
      return 1;
    else
      return 0;
  }
}
int main(int argc, char *argv[]) {
  int d, ap, bp, n;
  scanf("%d%s%s", &n, a, b);
  int i, j, k, al, bl, t;
  al = strlen(a);
  bl = strlen(b);
  d = al * bl / gcd(al, bl);
  ap = 0, bp = 0;
  for (k = 0, i = 0, j = 0; k < d; ++k, i = (i + 1) % al, j = (j + 1) % bl) {
    t = cmp(a[i], b[j]);
    if (t < 0)
      ++ap;
    else if (t > 0)
      ++bp;
  }
  int ar = n / d * ap, br = n / d * bp;
  int td = n % d;
  ap = 0, bp = 0;
  for (k = 0, i = 0, j = 0; k < td; ++k, i = (i + 1) % al, j = (j + 1) % bl) {
    t = cmp(a[i], b[j]);
    if (t < 0)
      ++ap;
    else if (t > 0)
      ++bp;
  }
  printf("%d %d\n", ar + ap, br + bp);
  return 0;
}
