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 cnt1[1000000];
int cnt2[1000000];
int res(char a, char b) {
  if (a == 'S') {
    if (b == 'S') return 0;
    if (b == 'R') return 1;
    return 2;
  }
  if (a == 'R') {
    if (b == 'R') return 0;
    if (b == 'P') return 1;
    return 2;
  }
  if (b == 'P') return 0;
  if (b == 'S') return 1;
  return 2;
}
int main() {
  int n;
  cin >> n;
  string s1, s2;
  cin >> s1 >> s2;
  int k = s1.length();
  int m = s2.length();
  int q;
  q = res(s1[0], s2[0]);
  if (q == 1) {
    cnt1[0] = 1;
    cnt2[0] = 0;
  } else if (q == 2) {
    cnt1[0] = 0;
    cnt2[0] = 1;
  } else {
    cnt1[0] = 0;
    cnt2[0] = 0;
  }
  for (int i = 1; i < k * m; i++) {
    q = res(s1[i % k], s2[i % m]);
    if (q == 1) {
      cnt1[i] = cnt1[i - 1] + 1;
      cnt2[i] = cnt2[i - 1];
    } else if (q == 2) {
      cnt1[i] = cnt1[i - 1];
      cnt2[i] = cnt2[i - 1] + 1;
    } else {
      cnt1[i] = cnt1[i - 1];
      cnt2[i] = cnt2[i - 1];
    }
  }
  int z = n / (k * m);
  int w = n % (k * m);
  {
    printf("%d %d", cnt1[k * m - 1] * z + (w > 0 ? cnt1[w - 1] : 0),
           cnt2[k * m - 1] * z + (w > 0 ? cnt2[w - 1] : 0));
  }
}
