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 == 0) return a;
  return gcd(b, a % b);
}
char alice[1100], bob[1100];
int tran[256];
void prework() {
  tran['R'] = 0;
  tran['S'] = 1;
  tran['P'] = 2;
}
int main() {
  int n;
  prework();
  while (scanf(" %d", &n) == 1) {
    int alice_red, bob_red;
    alice_red = bob_red = 0;
    scanf(" %s %s", alice, bob);
    int alice_len = strlen(alice);
    int bob_len = strlen(bob);
    int i, j;
    i = j = 0;
    do {
      if ((tran[(int)alice[i]] + 1) % 3 == tran[(int)bob[j]]) {
        bob_red++;
      }
      if ((tran[(int)bob[j]] + 1) % 3 == tran[(int)alice[i]]) {
        alice_red++;
      }
      i = (i + 1) % alice_len;
      j = (j + 1) % bob_len;
    } while (i | j);
    int round_c = alice_len / gcd(alice_len, bob_len) * bob_len;
    int play_round = n / round_c;
    int out_alice = alice_red * play_round;
    int out_bob = bob_red * play_round;
    n -= round_c * play_round;
    i = j = 0;
    for (int k = 0; k < n; k++) {
      if ((tran[(int)alice[i]] + 1) % 3 == tran[(int)bob[j]]) {
        out_bob++;
      }
      if ((tran[(int)bob[j]] + 1) % 3 == tran[(int)alice[i]]) {
        out_alice++;
      }
      i = (i + 1) % alice_len;
      j = (j + 1) % bob_len;
    }
    printf("%d %d\n", out_alice, out_bob);
    fflush(stdout);
  }
  return 0;
}
