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;
string str1, str2, tmp1, tmp2;
int ans[3][1000005];
int gcd(int x, int y) {
  if (!y) return x;
  return gcd(y, x % y);
}
int cal(char p, char q) {
  if (p == q) return 0;
  if ((p == 'R' && q == 'P') || (p == 'P' && q == 'S') ||
      (p == 'S' && q == 'R'))
    return 1;
  return 2;
}
int main() {
  int i, j, k, l1, l2, l, n;
  scanf("%d", &n);
  cin >> tmp1 >> tmp2;
  str1 = str2 = "";
  l1 = tmp1.length();
  l2 = tmp2.length();
  l = l1 * l2 / gcd(l1, l2);
  k = l / l1;
  for (i = 1; i <= k; i++) str1 += tmp1;
  k = l / l2;
  for (i = 1; i <= k; i++) str2 += tmp2;
  for (i = 1; i <= l; i++) {
    ans[cal(str1[i - 1], str2[i - 1])][i]++;
    ans[1][i] += ans[1][i - 1];
    ans[2][i] += ans[2][i - 1];
    if (i == n) break;
  }
  if (n <= l) {
    printf("%d %d\n", ans[1][n], ans[2][n]);
    return 0;
  }
  j = n / l;
  k = n % l;
  printf("%d %d\n", ans[1][l] * j + ans[1][k], ans[2][l] * j + ans[2][k]);
  return 0;
}
