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 <typename T>
T gcd(T a, T b) {
  return !b ? a : gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  return a * (b / gcd(a, b));
}
template <typename T>
T sqr(T a) {
  return a * a;
}
template <typename T>
T cube(T a) {
  return a * a * a;
}
template <typename T>
inline void smin(T &a, T b) {
  a = a < b ? a : b;
}
template <typename T>
inline void smax(T &a, T b) {
  a = a > b ? a : b;
}
template <typename T>
void fill(T a[], T n, T b) {
  for (int i = 0; i <= n; i++) a[i] = b;
}
int in() {
  int n;
  scanf("%d", &n);
  return n;
}
long long Lin() {
  long long n;
  scanf("%lld", &n);
  return n;
}
double Din() {
  double n;
  scanf("%lf", &n);
  return n;
}
const long long inf = (long long)1e17;
const long long mod = (long long)1e9 + 7;
const int N = 2e6 + 5;
int dp[N], dp2[N];
int ok(char a, char b) {
  if (a == 'R' && b == 'S') return 1;
  if (a == 'S' && b == 'P') return 1;
  if (a == 'P' && b == 'R') return 1;
  return 0;
}
int solve() {
  int n = in();
  string s, t, second = "", tt = "";
  cin >> s >> t;
  int a = (int)s.size(), b = (int)t.size();
  int l = lcm(a, b);
  while ((int)second.size() != l) second += s;
  while ((int)tt.size() != l) tt += t;
  for (int i = 1; i <= l; i++) {
    dp[i] += dp[i - 1];
    dp2[i] += dp2[i - 1];
    if (second[i - 1] == tt[i - 1]) continue;
    if (ok(second[i - 1], tt[i - 1]))
      dp2[i]++;
    else
      dp[i]++;
  }
  int ans = (n / l) * dp[l] + dp[n % l], ans2 = (n / l) * dp2[l] + dp2[n % l];
  cout << ans << " " << ans2 << '\n';
  return 0;
}
int main() {
  int test = 1, tc = 0;
  while (test--) {
    solve();
  }
  return 0;
}
