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;
const int MAXN = 1009;
const double eps = 1e-9;
int n, m, k;
vector<char> M;
vector<char> K;
int gcd(int a, int b) {
  if (!a) return b;
  return gcd(b % a, a);
}
int win(char a, char b) {
  if (a == 'R' && b == 'S') return 1;
  if (a == 'S' && b == 'P') return 1;
  if (a == 'P' && b == 'R') return 1;
  if (b == 'R' && a == 'S') return -1;
  if (b == 'S' && a == 'P') return -1;
  if (b == 'P' && a == 'R') return -1;
  return 0;
}
pair<int, int> emulate(int t) {
  int result1 = 0;
  int result2 = 0;
  for (int i = 0; i < t; i++) {
    int tmp = win(M[i % m], K[i % k]);
    if (tmp == 1) result1++;
    if (tmp == -1) result2++;
  }
  return make_pair(result1, result2);
}
int main() {
  scanf("%d\n", &n);
  char c = ' ';
  while (c != '\n') {
    scanf("%c", &c);
    if (c == '\n') break;
    M.push_back(c);
  }
  c = ' ';
  while (c != '\n') {
    scanf("%c", &c);
    if (c == '\n') break;
    K.push_back(c);
  }
  m = M.size();
  k = K.size();
  int g = m * k / gcd(m, k);
  pair<int, int> diff = emulate(g);
  pair<int, int> tmp = emulate(n % g);
  pair<int, int> ans(tmp.first + diff.first * (n / g),
                     tmp.second + diff.second * (n / g));
  printf("%d %d\n", ans.second, ans.first);
  return 0;
}
