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;
bool can[10][10];
char s[1010], t[1010];
inline int get(char ch) {
  if (ch == 'S') {
    return 0;
  }
  if (ch == 'P') {
    return 1;
  }
  return 2;
}
inline int gcd(int a, int b) {
  while (b > 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int main() {
  can[2][0] = can[0][1] = can[1][2] = true;
  int n;
  scanf("%d %s %s", &n, s, t);
  int m = strlen(s), k = strlen(t);
  int nok = (m * k) / gcd(m, k), add1 = 0, add2 = 0;
  for (int i = 0; i < nok; i++) {
    if (can[get(s[i % m])][get(t[i % k])]) {
      add2++;
    }
    if (can[get(t[i % k])][get(s[i % m])]) {
      add1++;
    }
  }
  int ans1 = add1 * (n / nok), ans2 = add2 * (n / nok);
  for (int i = 0; i < n % nok; i++) {
    if (can[get(s[i % m])][get(t[i % k])]) {
      ans2++;
    }
    if (can[get(t[i % k])][get(s[i % m])]) {
      ans1++;
    }
  }
  printf("%d %d\n", ans1, ans2);
  return 0;
}
