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) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
int Nik = 0, Pol = 0;
string s1;
string s2;
int n, bound;
char nikm, polm;
void f(int boun) {
  for (int i = 0; i < boun; i++) {
    nikm = s1[i % s1.size()];
    polm = s2[i % s2.size()];
    if (nikm == polm) continue;
    if (nikm == 'S' && polm == 'P') {
      Pol++;
      continue;
    }
    if (nikm == 'P' && polm == 'R') {
      Pol++;
      continue;
    }
    if (nikm == 'R' && polm == 'S') {
      Pol++;
      continue;
    } else
      Nik++;
  }
}
int main(void) {
  scanf("%d", &n);
  long long lcmNP;
  cin >> s1;
  cin >> s2;
  lcmNP = lcm(s1.size(), s2.size());
  if (!(n / lcmNP))
    bound = 0;
  else {
    bound = lcmNP;
    f(bound);
    Nik = Nik * (n / lcmNP);
    Pol = Pol * (n / lcmNP);
  }
  bound = n % lcmNP;
  f(bound);
  printf("%d %d\n", Nik, Pol);
  return 0;
}
