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;
const int INF = 2147483647;
const double PI = 3.141592653589793;
const int N = 1000005;
const int M = 1005;
int res1[N], res2[N], n1, n2, n, d1, d2, nw, i;
char t1[M], t2[M];
string s1, s2;
char c1, c2;
int nwd(int a, int b) {
  if (b == 0) return a;
  return nwd(b, a % b);
}
int nww(int a, int b) { return a * b / nwd(a, b); }
bool better(char c1, char c2) {
  if (c1 == 'R') return (c2 == 'S');
  if (c1 == 'S') return (c2 == 'P');
  if (c1 == 'P') return (c2 == 'R');
}
int main() {
  scanf("%d %s %s", &n, t1, t2);
  s1 = t1;
  s2 = t2;
  n1 = s1.size();
  n2 = s2.size();
  nw = nww(n1, n2);
  d1 = d2 = 0;
  for (i = 0; i < nw; i++) {
    c1 = s1[i % n1];
    c2 = s2[i % n2];
    if (better(c1, c2))
      d2++;
    else if (better(c2, c1))
      d1++;
    res1[i + 1] = d1;
    res2[i + 1] = d2;
  }
  printf("%d %d\n", res1[nw] * (n / nw) + res1[n % nw],
         res2[nw] * (n / nw) + res2[n % nw]);
  return 0;
}
