#include <bits/stdc++.h>
using namespace std;
inline bool feq(const double& a, const double& b) {
  return fabs(a - b) < 1e-10;
}
int N;
int has[10] = {0};
int main() {
  cin >> N;
  for (int i = 0; i < N; ++i) {
    int a;
    cin >> a;
    ++has[a / 400];
  }
  int cnt = 0;
  for (int i = 0; i < 8; ++i) cnt += bool(has[i]);
  int mi = cnt;
  int ma = min(cnt + has[8], 8);
  cout << mi << " " << ma << endl;
  return 0;
}
