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;
long long n, a, b, t1, t2, cnt1, cnt2, r2, r1;
char s1[1005], s2[1005];
int p1 = 0, p2 = 0, r = 0;
long long mcd(long long a, long long b) {
  if (b == 0) return a;
  return mcd(b, a % b);
}
void f(int tmp) {
  for (int i = 0; i < tmp; i++) {
    if ((s1[p1] == 'R' && s2[p2] == 'S') || (s1[p1] == 'S' && s2[p2] == 'P') ||
        (s1[p1] == 'P' && s2[p2] == 'R'))
      cnt1++;
    else if (s1[p1] != s2[p2])
      cnt2++;
    p1 = (p1 + 1) % t1;
    p2 = (p2 + 1) % t2;
  }
}
int main() {
  scanf("%I64d", &n);
  scanf("%s%s", s1, s2);
  p1 = 0, p2 = 0, t1 = strlen(s1), t2 = strlen(s2);
  long long tmp = t1 / (mcd(t1, t2)) * t2;
  if (n < tmp) {
    cnt1 = 0;
    cnt2 = 0;
    f(n);
    printf("%I64d %I64d\n", cnt2, cnt1);
  } else {
    r1 = n / tmp;
    cnt1 = 0;
    cnt2 = 0;
    f(tmp);
    r2 = r1 * cnt2;
    r1 = r1 * cnt1;
    cnt1 = 0;
    cnt2 = 0;
    f(n % tmp);
    r1 += cnt1;
    r2 += cnt2;
    printf("%I64d %I64d\n", r2, r1);
  }
  return 0;
}
