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;
struct BO {
  int a, b;
} a[300];
int n, tot, mt, tt;
int f[110][11000];
int cmp(const BO &x, const BO &y) { return x.b < y.b; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].a;
    tot += a[i].a;
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i].b;
    mt += a[i].b;
  }
  sort(a + 1, a + n + 1, cmp);
  int K = 0;
  tt = tot;
  for (int i = n; i >= 1 && tot > 0; i--) {
    K++;
    tot -= a[i].b;
  }
  int ans = 0;
  memset(f, -127 / 3, sizeof(f));
  f[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = K; j >= 1; j--)
      for (int k = mt; k >= a[i].b; k--)
        f[j][k] = max(f[j][k], f[j - 1][k - a[i].b] + a[i].a);
  }
  for (int k = tt; k <= mt; k++) ans = max(ans, f[K][k]);
  cout << K << ' ' << tt - ans << endl;
}
