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 n, cnt[1000005][2];
int ls, lt;
char s[1005], t[1005];
int res[3][3] = {{0, -1, 1}, {1, 0, -1}, {-1, 1, 0}};
int main() {
  scanf("%d\n%s\n%s", &n, s, t);
  ls = strlen(s);
  lt = strlen(t);
  memset(cnt, 0, sizeof(cnt));
  for (int i = (0); i <= (ls * lt - 1); i++) {
    if (i) cnt[i][0] = cnt[i - 1][0], cnt[i][1] = cnt[i - 1][1];
    int n1, n2;
    if (s[i % ls] == 'R')
      n1 = 0;
    else if (s[i % ls] == 'P')
      n1 = 1;
    else if (s[i % ls] == 'S')
      n1 = 2;
    if (t[i % lt] == 'R')
      n2 = 0;
    else if (t[i % lt] == 'P')
      n2 = 1;
    else if (t[i % lt] == 'S')
      n2 = 2;
    if (res[n1][n2] == -1) cnt[i][0]++;
    if (res[n1][n2] == 1) cnt[i][1]++;
  }
  for (int i = (ls * lt); i >= (1); i--)
    cnt[i][0] = cnt[i - 1][0], cnt[i][1] = cnt[i - 1][1];
  cnt[0][0] = cnt[0][1] = 0;
  printf("%d %d\n", n / (ls * lt) * cnt[ls * lt][0] + cnt[n % (ls * lt)][0],
         n / (ls * lt) * cnt[ls * lt][1] + cnt[n % (ls * lt)][1]);
  return 0;
}
