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;
template <class T>
inline bool Up(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool Down(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
int N, l, l1, l2;
int sum1[1024 * 1024], sum2[1024 * 1024], beat[256][256];
char s1[1024], s2[1024];
inline int Gcd(int a, int b) {
  for (int c; c = a % b; a = b, b = c)
    ;
  return b;
}
inline int Lcm(int a, int b) { return a / Gcd(a, b) * b; }
void Work() {
  l = Lcm(l1, l2);
  for (int i = (0); i < (l); ++i) {
    int a = s1[i % l1], b = s2[i % l2];
    sum1[i] = sum2[i] = 0;
    if (beat[a][b])
      sum1[i] = 0, sum2[i] = 1;
    else if (beat[b][a])
      sum1[i] = 1, sum2[i] = 0;
    else
      sum1[i] = sum2[i] = 0;
    if (i) {
      sum1[i] += sum1[i - 1];
      sum2[i] += sum2[i - 1];
    }
  }
  int x, y;
  x = sum1[l - 1] * (N / l);
  y = sum1[N % l - 1];
  printf("%d ", x + y);
  x = sum2[l - 1] * (N / l);
  y = sum2[N % l - 1];
  printf("%d\n", x + y);
}
void Init() {
  memset(beat, 0, sizeof(beat));
  scanf("%d", &N);
  scanf("%s", s1);
  scanf("%s", s2);
  l1 = strlen(s1);
  l2 = strlen(s2);
  beat['R']['S'] = 1;
  beat['S']['P'] = 1;
  beat['P']['R'] = 1;
}
int main() {
  Init();
  Work();
  return 0;
}
