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>
#pragma warning(disable : 4996)
using namespace std;
char w[128][128], a[1002], b[1002];
int main() {
  int n, m, r, i, j, k, w1 = 0, w2 = 0, i1, i2, repeat;
  scanf("%d", &r);
  scanf("%s%s", a, b);
  n = strlen(a);
  m = strlen(b);
  i1 = i2 = 0;
  w['R']['S'] = 1;
  w['S']['P'] = 1;
  w['P']['R'] = 1;
  for (i = 0; i < n * m; ++i) {
    if (a[i1] != b[i2])
      if (w[a[i1]][b[i2]])
        ++w1;
      else
        ++w2;
    i1 = (i1 + 1) % n;
    i2 = (i2 + 1) % m;
  }
  repeat = r / (n * m);
  w1 *= repeat;
  w2 *= repeat;
  r %= n * m;
  for (i = 0; i < r; ++i) {
    if (a[i1] != b[i2])
      if (w[a[i1]][b[i2]])
        ++w1;
      else
        ++w2;
    i1 = (i1 + 1) % n;
    i2 = (i2 + 1) % m;
  }
  printf("%d ", w2);
  printf("%d ", w1);
  return 0;
}
