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 T1>
void deb(T1 e) {
  cout << e << endl;
}
template <class T1, class T2>
void deb(T1 e1, T2 e2) {
  cout << e1 << " " << e2 << endl;
}
template <class T1, class T2, class T3>
void deb(T1 e1, T2 e2, T3 e3) {
  cout << e1 << " " << e2 << " " << e3 << endl;
}
template <class T1, class T2, class T3, class T4>
void deb(T1 e1, T2 e2, T3 e3, T4 e4) {
  cout << e1 << " " << e2 << " " << e3 << " " << e4 << endl;
}
template <class T1, class T2, class T3, class T4, class T5>
void deb(T1 e1, T2 e2, T3 e3, T4 e4, T5 e5) {
  cout << e1 << " " << e2 << " " << e3 << " " << e4 << " " << e5 << endl;
}
template <class T1, class T2, class T3, class T4, class T5, class T6>
void deb(T1 e1, T2 e2, T3 e3, T4 e4, T5 e5, T6 e6) {
  cout << e1 << " " << e2 << " " << e3 << " " << e4 << " " << e5 << " " << e6
       << endl;
}
long long sqr(long long n) { return n * n; }
double LOG(long long a, long long b) { return (log(a) / log(b)); }
long long GCD(long long a, long long b) {
  if (a < 0) return GCD(-a, b);
  if (b < 0) return GCD(a, -b);
  return (!b) ? a : GCD(b, a % b);
}
long long LCM(long long a, long long b) {
  if (a < 0) return LCM(-a, b);
  if (b < 0) return LCM(a, -b);
  return (a * (b / GCD(a, b)));
}
long long BigMod(long long B, long long P, long long M) {
  if (!P) return 1;
  if (!(P % 2)) return sqr(BigMod(B, P / 2, M)) % M;
  return (B % M) * (BigMod(B, P - 1, M)) % M;
}
int rN = 0, rP = 0;
string a, b;
int tag, sA, sB;
void process(int i) {
  char cN, cR;
  if (!(i % sA))
    cN = a[sA - 1];
  else
    cN = a[(i % sA) - 1];
  if (!(i % sB))
    cR = b[sB - 1];
  else
    cR = b[(i % sB) - 1];
  if (cN == cR) return;
  if (cN == 'R') {
    if (cR == 'S')
      rP++;
    else
      rN++;
  } else if (cN == 'S') {
    if (cR == 'P')
      rP++;
    else
      rN++;
  } else if (cN == 'P') {
    if (cR == 'R')
      rP++;
    else
      rN++;
  }
  return;
}
int main(void) {
  int n;
  scanf("%d", &n);
  cin >> a >> b;
  tag = (int)a.size() * (int)b.size();
  sA = (int)a.size(), sB = (int)b.size();
  int i;
  if (tag >= n) {
    for (i = 1; i <= n; i++) process(i);
    deb(rN, rP);
    return 0;
  }
  for (i = 1; i <= tag; i++) process(i);
  int tmp = n / tag;
  rP *= tmp;
  rN *= tmp;
  for (i = tmp * tag + 1; i <= n; i++) process(i);
  deb(rN, rP);
  return 0;
}
