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 n, m;
string str1, str2;
int ans1, ans2;
int who(char a, char b) {
  if ((a == 'P' && b == 'S') || (a == 'S' && b == 'R') ||
      (a == 'R' && b == 'P'))
    return 1;
  if ((a == 'S' && b == 'P') || (a == 'R' && b == 'S') ||
      (a == 'P' && b == 'R'))
    return -1;
  return 0;
}
void fun1() {
  int i = 0, j = 0;
  while (true) {
    int w = who(str1[i], str2[j]);
    if (w == 1) {
      ans1++;
    } else if (w == -1)
      ans2++;
    i++, j++, m++;
    if (i == str1.size()) i = 0;
    if (j == str2.size()) j = 0;
    if (i == 0 && j == 0) {
      break;
    }
  }
}
void fun2() {
  int mod = n % m, x = n / m;
  ans1 *= x;
  ans2 *= x;
  int i = 0, j = 0;
  while (mod--) {
    int w = who(str1[i], str2[j]);
    if (w == 1) {
      ans1++;
    } else if (w == -1)
      ans2++;
    i++, j++;
    if (i == str1.size()) i = 0;
    if (j == str2.size()) j = 0;
  }
}
int main(int argc, char** argv) {
  while (cin >> n) {
    ans1 = ans2 = m = 0;
    cin >> str1 >> str2;
    fun1();
    fun2();
    cout << ans1 << " " << ans2 << endl;
  }
  return 0;
}
