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>
char a[1005], b[1005];
int alen, blen, cnt1, cnt2;
int gcd(int a, int b) {
  if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
int lcm(int a, int b) { return a * b / gcd(a, b); }
void work(int t) {
  int i;
  for (i = 0; i < t; i++) switch (a[i % alen] - b[i % blen]) {
      case 1:
        cnt1++;
        break;
      case 2:
        cnt1++;
        break;
      case 3:
        cnt2++;
        break;
      case -1:
        cnt2++;
        break;
      case -2:
        cnt2++;
        break;
      case -3:
        cnt1++;
        break;
    }
}
int main() {
  int i, n, t, s;
  while (~scanf("%d", &n)) {
    cnt1 = 0, cnt2 = 0;
    scanf("%s%s", a, b);
    alen = strlen(a);
    blen = strlen(b);
    t = lcm(alen, blen);
    work(t);
    s = n / t;
    cnt1 *= s;
    cnt2 *= s;
    work(n % t);
    printf("%d %d\n", cnt1, cnt2);
  }
  return 0;
}
