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;
string s1, s2;
int x, y;
int g(int a, int b) {
  if (a < b) return g(b, a);
  if (a % b == 0) {
    return b;
  } else
    return g(b, a % b);
}
void h(int m, int k, string a1, string a2) {
  for (int i = m; i < k; i++) {
    if (a1[i] == a2[i]) continue;
    if (a1[i] == 'R' && a2[i] == 'S') x++;
    if (a1[i] == 'R' && a2[i] == 'P') y++;
    if (a1[i] == 'S' && a2[i] == 'P') x++;
    if (a1[i] == 'S' && a2[i] == 'R') y++;
    if (a1[i] == 'P' && a2[i] == 'R') x++;
    if (a1[i] == 'P' && a2[i] == 'S') y++;
  }
}
int main() {
  int n;
  while (cin >> n) {
    string a1, a2;
    x = 0;
    y = 0;
    s1 = "";
    s2 = "";
    if (n < 1) break;
    cin >> s1;
    a1 = s1;
    cin >> s2;
    a2 = s2;
    int z1 = s1.length();
    int z2 = s2.length();
    int z = z1 * z2 / g(z1, z2);
    for (int i = 1; i < z / z1; i++) {
      s1 += a1;
    }
    for (int i = 1; i < z / z2; i++) {
      s2 += a2;
    }
    int left = n % z;
    n = n / z;
    h(0, z, s1, s2);
    x = x * n;
    y = y * n;
    h(0, left, s1, s2);
    cout << y << " " << x;
  }
  return 0;
}
