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 str1[1024], str2[1024];
unsigned int sum1 = 0, sum2 = 0;
unsigned int n, len1, len2;
unsigned int GetLcm(unsigned, unsigned);
void Compute(unsigned int);
bool FirstWin(int, int);
bool SecondWin(int, int);
int main() {
  cin >> n;
  cin >> str1 >> str2;
  len1 = strlen(str1);
  len2 = strlen(str2);
  unsigned int lcm = GetLcm(len1, len2);
  if (n >= lcm) Compute(lcm);
  sum1 *= (n / lcm);
  sum2 *= (n / lcm);
  Compute(n % lcm);
  cout << sum1 << ' ' << sum2;
  return 0;
}
unsigned int GetLcm(unsigned int x, unsigned int y) {
  unsigned int mul = x * y;
  unsigned int t;
  if (x < y) {
    t = x;
    x = y;
    y = t;
  }
  t = x % y;
  while (t) {
    x = y;
    y = t;
    t = x % y;
  }
  return (mul / y);
}
void Compute(unsigned int round) {
  unsigned int i;
  int k = 0, j = 0;
  for (i = 1; i <= round; ++i) {
    if (FirstWin(j, k))
      sum2++;
    else if (SecondWin(j, k))
      sum1++;
    j = (j + 1) % len1;
    k = (k + 1) % len2;
  }
}
bool FirstWin(int j, int k) {
  return (str1[j] == 'P' && str2[k] == 'R' ||
          str1[j] == 'R' && str2[k] == 'S' || str1[j] == 'S' && str2[k] == 'P');
}
bool SecondWin(int j, int k) {
  return (str1[j] == 'P' && str2[k] == 'S' ||
          str1[j] == 'R' && str2[k] == 'P' || str1[j] == 'S' && str2[k] == 'R');
}
