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 gcd(int a, int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  while (b > 0) {
    int r = a % b;
    a = b;
    b = r;
  }
  return a;
}
int mcm(int a, int b) { return (long long)a * b / gcd(a, b); }
int winner(char a, char b) {
  if (a == b)
    return 0;
  else if (a == 'R') {
    if (b == 'S')
      return 1;
    else
      return 2;
  } else if (a == 'S') {
    if (b == 'P')
      return 1;
    else
      return 2;
  } else {
    if (b == 'R')
      return 1;
    else
      return 2;
  }
}
int main() {
  int n;
  cin >> n;
  string A, B;
  cin >> A >> B;
  string nA, nB;
  int m = mcm(A.size(), B.size());
  for (int i = 0; i < m / A.size(); i++) {
    nA.append(A);
  }
  for (int i = 0; i < m / B.size(); i++) {
    nB.append(B);
  }
  int wsf1[m], wsf2[m];
  int totA = 0, totB = 0;
  if (winner(nA[0], nB[0]) == 0)
    wsf1[0] = wsf2[0] = 0;
  else if (winner(nA[0], nB[0]) == 1)
    wsf1[0] = 1, wsf2[0] = 0, totA++;
  else
    wsf1[0] = 0, wsf2[0] = 1, totB++;
  for (int i = 1; i < m; i++) {
    if (winner(nA[i], nB[i]) == 0)
      wsf1[i] = wsf1[i - 1], wsf2[i] = wsf2[i - 1];
    else if (winner(nA[i], nB[i]) == 1)
      wsf1[i] = wsf1[i - 1] + 1, wsf2[i] = wsf2[i - 1], totA++;
    else
      wsf1[i] = wsf1[i - 1], wsf2[i] = wsf2[i - 1] + 1, totB++;
  }
  int a = n / m * totA;
  int b = n / m * totB;
  if (n % m) {
    a += wsf1[n % m - 1];
    b += wsf2[n % m - 1];
  }
  cout << b << " " << a;
}
