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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  string ns, ps;
  map<int, pair<int, int>> hg;
  int n, nn, pn, nc = 0, pc = 0, mn, ni = 0, pi = 0, M;
  cin >> n >> ns >> ps;
  nn = ns.size();
  pn = ps.size();
  mn = pn * nn;
  M = mn;
  while (mn > 0) {
    if (ns[ni] != ps[pi]) {
      if ((ns[ni] == 'R' && ps[pi] == 'S') ||
          (ns[ni] == 'P' && ps[pi] == 'R') || (ns[ni] == 'S' && ps[pi] == 'P'))
        pc += 1;
      else
        nc += 1;
    }
    hg[M - mn].first = nc;
    hg[M - mn].second = pc;
    ni = (ni + 1) % nn;
    pi = (pi + 1) % pn;
    mn--;
  }
  if (n <= M)
    cout << hg[n - 1].first << " " << hg[n - 1].second;
  else {
    int q = n / M;
    n %= M;
    cout << hg[M - 1].first * q + hg[n - 1].first << " "
         << hg[M - 1].second * q + hg[n - 1].second;
  }
  return 0;
}
