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, f[110][110 * 110];
pair<int, int> x[110];
bool cmp(pair<int, int> a, pair<int, int> b) { return a.second > b.second; }
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  int sum = 0;
  for (int i = 1; i <= n; i++) cin >> x[i].first, sum += x[i].first;
  for (int i = 1; i <= n; i++) cin >> x[i].second;
  sort(x + 1, x + n + 1, cmp);
  int vt = 0, temp = 0;
  for (int i = 1; i <= n; i++) {
    temp += x[i].second;
    if (temp >= sum) {
      vt = i;
      break;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = min(i - 1, vt); j >= 1; j--)
      for (int k = sum + 1; k >= 0; k--) {
        if (f[j][k] == 0) continue;
        if (k + x[i].second > sum) {
          f[j + 1][sum + 1] = max(f[j + 1][sum + 1], f[j][k] + x[i].first);
        } else
          f[j + 1][k + x[i].second] =
              max(f[j + 1][k + x[i].second], f[j][k] + x[i].first);
      }
    f[1][min(sum + 1, x[i].second)] =
        max(f[1][min(sum + 1, x[i].second)], x[i].first);
  }
  cout << vt << " ";
  cout << sum - max(f[vt][sum], f[vt][sum + 1]);
}
