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:64000000")
using namespace std;
const long long INF = ((1 << 31) - 1);
const long long LLINF = (((1LL << 63) - 1LL));
const double EPS = 1e-9;
const double PI = 3.14159265358979323846;
pair<long long, long long> operator+(const pair<long long, long long>& lhs,
                                     const pair<long long, long long>& rhs) {
  return pair<long long, long long>(lhs.first + rhs.first,
                                    lhs.second + rhs.second);
}
pair<long long, long long> play(char a, char b) {
  if (a == b) {
    return pair<long long, long long>(0, 0);
  }
  if (a == 'R') {
    if (b == 'S') {
      return pair<long long, long long>(0, 1);
    } else {
      return pair<long long, long long>(1, 0);
    }
  }
  if (a == 'S') {
    if (b == 'P') {
      return pair<long long, long long>(0, 1);
    } else {
      return pair<long long, long long>(1, 0);
    }
  }
  if (a == 'P') {
    if (b == 'R') {
      return pair<long long, long long>(0, 1);
    } else {
      return pair<long long, long long>(1, 0);
    }
  }
  assert(false);
}
pair<long long, long long> solve(const string& a, const string& b, int n) {
  int M = a.size() * b.size();
  string aa;
  for (int i = 0; i < M / a.size(); ++i) {
    aa += a;
  }
  string bb;
  for (int i = 0; i < M / b.size(); ++i) {
    bb += b;
  }
  long long times = n / M;
  long long pred = n % M;
  pair<long long, long long> res(0, 0);
  for (int i = 0; i < pred; ++i) {
    pair<long long, long long> add = play(aa[i], bb[i]);
    res = res + add;
  }
  pair<long long, long long> perPer(0, 0);
  for (int i = 0; i < M; ++i) {
    pair<long long, long long> add = play(aa[i], bb[i]);
    perPer = perPer + add;
  }
  res.first += perPer.first * times;
  res.second += perPer.second * times;
  return res;
}
int main() {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  pair<long long, long long> res = solve(a, b, n);
  cout << res.first << " " << res.second << endl;
  return 0;
}
