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 s1[1010], s2[1010];
int gao(char c1, char c2) {
  if (c1 == c2) return 0;
  if (c1 == 'R' && c2 == 'S' || c1 == 'S' && c2 == 'P' ||
      c1 == 'P' && c2 == 'R') {
    return 1;
  }
  return -1;
}
int main(void) {
  int ans1 = 0, ans2 = 0;
  int n;
  scanf("%d", &n);
  scanf("%s", s1);
  scanf("%s", s2);
  int len1 = strlen(s1);
  int len2 = strlen(s2);
  int sum = len1 / gcd(len1, len2) * len2;
  int tmp1 = 0, tmp2 = 0, t = 0;
  if (sum < n) {
    t = n % sum;
    for (int i = 0; i < sum; i++) {
      int u = i % len1, v = i % len2;
      if (gao(s1[u], s2[v]) == 1) {
        ans1++;
        if (i < t) tmp1++;
      } else if (gao(s1[u], s2[v]) == -1) {
        ans2++;
        if (i < t) tmp2++;
      }
    }
    ans1 *= (n / sum);
    ans2 *= (n / sum);
    ans1 += tmp1;
    ans2 += tmp2;
  } else {
    for (int i = 0; i < n; i++) {
      int u = i % len1, v = i % len2;
      if (gao(s1[u], s2[v]) == 1) {
        ans1++;
      } else if (gao(s1[u], s2[v]) == -1) {
        ans2++;
      }
    }
  }
  printf("%d %d\n", ans2, ans1);
  return 0;
}
