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;
const int maxn = 1000005;
int ans, cnt, n, step;
char first[maxn], last[maxn];
void check(char x, char y) {
  if (x == y) return;
  if (x == 'R' && y == 'P') {
    ans++;
    return;
  }
  if (x == 'R' && y == 'S') {
    cnt++;
    return;
  }
  if (x == 'P' && y == 'R') {
    cnt++;
    return;
  }
  if (x == 'P' && y == 'S') {
    ans++;
    return;
  }
  if (x == 'S' && y == 'R') {
    ans++;
    return;
  }
  if (x == 'S' && y == 'P') {
    cnt++;
    return;
  }
}
int gcd(int x, int y) {
  int maxn = max(x, y);
  for (int i = maxn; i <= x * y; i += maxn)
    if (i % x == 0 && i % y == 0) return i;
}
inline int read() {
  int ret = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (isdigit(ch)) ret = ret * 10 + ch - '0', ch = getchar();
  return ret * f;
}
int main() {
  n = read();
  scanf("%s", first + 1);
  scanf("%s", last + 1);
  int len = strlen(first + 1), len2 = strlen(last + 1);
  int m = gcd(len, len2);
  for (int i = len + 1; i <= m; i++)
    first[i] = first[++step], step == len ? step = 0 : step;
  step = 0;
  for (int i = len2 + 1; i <= m; i++)
    last[i] = last[++step], step == len2 ? step = 0 : step;
  for (int i = 1; i <= m; i++) check(first[i], last[i]);
  int size = n / m;
  n %= m;
  ans *= size;
  cnt *= size;
  for (int i = 1; i <= n; i++) check(first[i], last[i]);
  printf("%d %d\n", ans, cnt);
  return 0;
}
