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;
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int lcm(int a, int b) { return a / gcd(a, b) * b; }
int n, r1, r2, g;
string x, y;
int play(int X, int Y) {
  if (x[X] == y[Y]) return 0;
  if ((x[X] == 'S' && y[Y] == 'R') || (x[X] == 'R' && y[Y] == 'P') ||
      (x[X] == 'P' && y[Y] == 'S'))
    return -1;
  return 1;
}
void suml(int t) {
  int A = 0, B = 0;
  int p1 = 0, p2 = 0, i = 0;
  while (i < g) {
    ++i;
    int get = play(p1 % x.size(), p2 % y.size());
    ++p1;
    ++p2;
    A += get == 1;
    B += get == -1;
  }
  r1 += A * t;
  r2 += B * t;
}
int main() {
  cin >> n >> x >> y;
  g = lcm(x.size(), y.size());
  if (n >= g) {
    int t = n / g;
    suml(t);
    n -= t * g;
  }
  int p1 = 0, p2 = 0;
  while (n--) {
    int get = play(p1 % x.size(), p2 % y.size());
    if (get == -1)
      ++r2;
    else if (get == 1)
      ++r1;
    ++p1;
    ++p2;
  }
  cout << r2 << " " << r1 << endl;
}
