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>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
int a, b, c, d, i, j, n, m, k;
char s1[1002], s2[1002];
inline int gcd(int a, int b) {
  for (; b; swap(a, b)) {
    a %= b;
  }
  return a;
}
inline bool win(char a, char b) {
  if (a == 'R' & b == 'S') {
    return 1;
  } else if (a == 'S' && b == 'P') {
    return 1;
  } else if (a == 'P' && b == 'R') {
    return 1;
  } else {
    return 0;
  }
}
inline pair<int, int> solve(int n) {
  int a1 = 0, a2 = 0;
  for (int _n((n)-1), i(0); i <= _n; i++) {
    char c1 = s1[i % a];
    char c2 = s2[i % b];
    if (c1 == c2) {
      continue;
    }
    if (win(c1, c2)) {
      ++a1;
    } else {
      ++a2;
    }
  }
  return make_pair(a1, a2);
}
int main() {
  scanf("%d", &n);
  scanf("%s%s", s1, s2);
  a = (int)strlen(s1), b = (int)strlen(s2);
  c = a / gcd(a, b) * b;
  pair<int, int> t = solve(c);
  t.first *= (n / c);
  t.second *= (n / c);
  pair<int, int> q = solve(n % c);
  t.first += q.first;
  t.second += q.second;
  printf("%d %d\n", t.second, t.first);
  return 0;
}
