Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volume bi (ai ≤ bi).

Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends x seconds to pour x units of soda from one bottle to another.

Nick asks you to help him to determine k — the minimal number of bottles to store all remaining soda and t — the minimal time to pour soda into k bottles. A bottle can't store more soda than its volume. All remaining soda should be saved.

Input

The first line contains positive integer n (1 ≤ n ≤ 100) — the number of bottles.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the amount of soda remaining in the i-th bottle.

The third line contains n positive integers b1, b2, ..., bn (1 ≤ bi ≤ 100), where bi is the volume of the i-th bottle.

It is guaranteed that ai ≤ bi for any i.

Output

The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.

Examples

Input

4
3 3 4 3
4 7 6 5


Output

2 6


Input

2
1 1
100 100


Output

1 1


Input

5
10 30 5 6 24
10 41 7 8 24


Output

3 11

Note

In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it.
#include <bits/stdc++.h>
using namespace std;
int n, p, S1[1000000], S2[1000000], s1, s2, t, T;
bool M[1000000];
struct par {
  int v, t;
  bool operator<(const par &R) const {
    if (t != R.t) return t > R.t;
    return v > R.v;
  }
} A[100000];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i].v;
    t += A[i].v;
  }
  for (int i = 1; i <= n; i++) {
    cin >> A[i].t;
    T += A[i].t;
  }
  M[0] = true;
  fill(S1 + 1, S1 + T + 1, 1 << 30);
  for (int i = 1; i <= n; i++) {
    for (int j = T - A[i].t; j >= 0; j--) {
      if (M[j]) {
        M[j + A[i].t] = true;
        if (S1[j + A[i].t] > S1[j] + 1) {
          S1[j + A[i].t] = S1[j] + 1;
          S2[j + A[i].t] = S2[j] + A[i].v;
        }
        if (S1[j + A[i].t] == S1[j] + 1)
          S2[j + A[i].t] = max(S2[j + A[i].t], S2[j] + A[i].v);
      }
    }
  }
  s1 = 1 << 30;
  for (int i = t; i <= T; i++) {
    if (!M[i]) continue;
    if (s1 > S1[i]) {
      s1 = S1[i];
      s2 = S2[i];
    }
    if (s1 == S1[i]) s2 = max(s2, S2[i]);
  }
  cout << s1 << " " << t - s2 << "\n";
  return 0;
}
