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 comment(linker, "/stack:64000000")
using namespace std;
template <typename first>
inline first abs(const first& a) {
  return a < 0 ? -a : a;
}
template <typename first>
inline first sqr(const first& a) {
  return a * a;
}
const long double PI = 3.1415926535897932384626433832795;
const long double EPS = 1e-9;
const int INF = 1000 * 1000 * 1000;
const int N = 1000 * 1000 + 3;
int n1, n2, n;
char s1[N], s2[N];
int com;
int gcd(int a, int b) { return a == 0 ? b : gcd(b % a, a); }
bool bit(char c1, char c2) {
  if (c1 == 'R' && c2 == 'S') return true;
  if (c1 == 'S' && c2 == 'P') return true;
  if (c1 == 'P' && c2 == 'R') return true;
  return false;
}
pair<int, int> calc() {
  pair<int, int> ans = make_pair((0), (0));
  for (int i = 0; i < int(com); i++)
    if (bit(s1[i], s2[i]))
      ans.second++;
    else if (bit(s2[i], s1[i]))
      ans.first++;
  return ans;
}
int main() {
  scanf("%d\n", &n);
  gets(s1);
  gets(s2);
  n1 = strlen(s1);
  n2 = strlen(s2);
  com = n1 / gcd(n1, n2) * n2;
  for (int i = 0; i < int(com); i++) {
    s1[i] = s1[i % n1];
    s2[i] = s2[i % n2];
  }
  int cntCom = n / com;
  pair<int, int> ans = calc();
  ans.first *= cntCom;
  ans.second *= cntCom;
  for (int i = 0; i < int(n % com); i++)
    if (bit(s1[i], s2[i]))
      ans.second++;
    else if (bit(s2[i], s1[i]))
      ans.first++;
  printf("%d %d\n", ans.first, ans.second);
  return 0;
}
