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:65000000")
using namespace std;
long long gcd(long long a, long long b) {
  while (a && b) {
    a %= b;
    swap(a, b);
  }
  return a + b;
}
long long lcd(long long a, long long b) { return a / gcd(a, b) * b; }
int n;
char a[1005], b[1005];
int m, k;
int mas[3][3] = {{0, 1, -1}, {-1, 0, 1}, {1, -1, 0}};
int gett(int pos) {
  char A = a[pos % m];
  char B = b[pos % k];
  return mas[A - 'Q'][B - 'Q'];
}
int main() {
  scanf("%d", &n);
  scanf(" %s %s", a, b);
  m = strlen(a);
  k = strlen(b);
  for (int i = 0; a[i]; ++i)
    if (a[i] == 'P') ++a[i];
  for (int i = 0; b[i]; ++i)
    if (b[i] == 'P') ++b[i];
  long long c = lcd(m, k);
  long long sum1 = 0;
  long long sum2 = 0;
  for (int i = 0; i < c; ++i) {
    int r = gett(i);
    sum1 += max(0, r);
    sum2 += -min(0, r);
  }
  sum1 *= n / c;
  sum2 *= n / c;
  c = n % c;
  for (int i = 0; i < c; ++i) {
    int r = gett(i);
    sum1 += max(0, r);
    sum2 += -min(0, r);
  }
  cout << sum2 << " " << sum1 << endl;
  return 0;
}
