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;
char s1[1001];
char s2[1001];
struct node {
  int a1, a2;
};
int n1, n2;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
node f(int m) {
  int m1, m2, a1, a2, i;
  node x;
  a1 = a2 = 0;
  for (i = 1; i <= m; i++) {
    if (i % n1 == 0)
      m1 = n1 - 1;
    else
      m1 = i % n1 - 1;
    if (i % n2 == 0)
      m2 = n2 - 1;
    else
      m2 = i % n2 - 1;
    if (s1[m1] == s2[m2]) continue;
    if (s1[m1] == 'R') {
      if (s2[m2] == 'P')
        a1++;
      else
        a2++;
    } else if (s1[m1] == 'P') {
      if (s2[m2] == 'R')
        a2++;
      else
        a1++;
    } else {
      if (s2[m2] == 'P')
        a2++;
      else
        a1++;
    }
  }
  x.a1 = a1;
  x.a2 = a2;
  return x;
}
int main() {
  int n, i, a1, a2, m, x, y;
  node A, B;
  while (~scanf("%d", &n)) {
    a1 = a2 = 0;
    scanf("%s%s", s1, s2);
    n1 = strlen(s1);
    n2 = strlen(s2);
    m = gcd(n1, n2);
    m = n1 / m * n2;
    x = n / m;
    y = n % m;
    A = f(y);
    if (x != 0) B = f(m);
    a1 = A.a1 + B.a1 * x;
    a2 = A.a2 + B.a2 * x;
    printf("%d %d\n", a1, a2);
  }
  return 0;
}
