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>
#pragma comment(linker, "/STACK:214721677")
using namespace std;
template <class T>
T sqr(T a) {
  return (a) * (a);
}
template <class T>
T abs(T a) {
  return (a < 0) ? -(a) : (a);
}
const double Pi = acos(-1.0);
const double eps = 1e-10;
const int INF = 1000 * 1000 * 1000;
const double phi = 0.5 + sqrt(1.25);
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  int n, m, k;
  string a, b;
  scanf("%d\n", &n);
  getline(cin, a);
  getline(cin, b);
  m = ((int)((a).size()));
  k = ((int)((b).size()));
  int len = gcd(m, k), lcm = m * k / len;
  string x = "", y = "";
  for (int i = 0, _n = (lcm / m) - 1; i <= _n; ++i) x += a;
  for (int i = 0, _n = (lcm / k) - 1; i <= _n; ++i) y += b;
  int countx = 0, county = 0;
  for (int i = 0, _n = (((int)((x).size()))) - 1; i <= _n; ++i) {
    countx += ((x[i] == 'S' && y[i] == 'P') || (x[i] == 'P' && y[i] == 'R') ||
               (x[i] == 'R' && y[i] == 'S'));
    county += ((y[i] == 'S' && x[i] == 'P') || (y[i] == 'P' && x[i] == 'R') ||
               (y[i] == 'R' && x[i] == 'S'));
  }
  int ansx = n / lcm * countx, ansy = n / lcm * county;
  n %= lcm;
  for (int i = 0; i < n; ++i) {
    ansx += ((x[i] == 'S' && y[i] == 'P') || (x[i] == 'P' && y[i] == 'R') ||
             (x[i] == 'R' && y[i] == 'S'));
    ansy += ((y[i] == 'S' && x[i] == 'P') || (y[i] == 'P' && x[i] == 'R') ||
             (y[i] == 'R' && x[i] == 'S'));
  }
  printf("%d %d\n", ansy, ansx);
  return 0;
}
