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 n;
long long gcd(long long x, long long y) {
  if (!x || !y) return max(x, y);
  for (long long t; t = x % y; x = y, y = t)
    ;
  return y;
}
int jugec(char a, char b) {
  if ((a == 'R' && b == 'P') || (a == 'P' && b == 'S') ||
      (a == 'S' && b == 'R'))
    return 1;
  else if (a == b)
    return -1;
  return 0;
}
void juge(char *st, char *sr) {
  int a = (long long)strlen(st);
  int b = (long long)strlen(sr);
  long long mid = (long long)a * b / gcd((long long)a, (long long)b);
  int sm1 = 0;
  int sm2 = 0;
  int i = 0;
  int j = 0;
  long long t = n / mid;
  long long tt = n % mid;
  if (t != 0) {
    for (long long k = 0; k < mid; k++) {
      if (jugec(st[i], sr[j]) == 1)
        sm1++;
      else if (jugec(st[i], sr[j]) == 0)
        sm2++;
      i++;
      j++;
      if (i == a) i = 0;
      if (j == b) j = 0;
    }
    sm1 = sm1 * t;
    sm2 = sm2 * t;
  }
  for (long long k = 0, i = 0, j = 0; k < tt; k++) {
    if (jugec(st[i], sr[j]) == 1)
      sm1++;
    else if (jugec(st[i], sr[j]) == 0)
      sm2++;
    i++;
    j++;
    if (i == a) i = 0;
    if (j == b) j = 0;
  }
  printf("%d %d\n", sm1, sm2);
}
int main() {
  char sr[1005], st[1005];
  while (scanf("%I64d", &n) != EOF) {
    scanf("%s", sr);
    scanf("%s", st);
    juge(sr, st);
  }
}
