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;
const int inf = INT_MAX;
const unsigned long long INF = LLONG_MAX;
const unsigned long long mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  unsigned long long t1 = 0, t2 = 0;
  for (int i = 0; i <= n - 1; i++) {
    cin >> a[i];
    t1 += a[i];
  }
  vector<int> b(n);
  for (int i = 0; i <= n - 1; i++) {
    cin >> b[i];
    t2 += b[i];
  }
  vector<long long int> vfilled(10001, -INF);
  vector<long long int> num(10001, INF);
  num[0] = 0;
  vfilled[0] = 0;
  for (int i = 0; i <= n - 1; i++) {
    for (int j = 10000 - b[i]; j >= 0; j--) {
      if (num[j] != INF) {
        if (num[j + b[i]] > num[j] + 1) {
          num[j + b[i]] = num[j] + 1;
          vfilled[j + b[i]] = vfilled[j] + a[i];
        } else if (num[j + b[i]] == num[j] + 1) {
          vfilled[j + b[i]] = max(vfilled[j + b[i]], vfilled[j] + a[i]);
        }
      }
    }
  }
  unsigned long long count = INF, ans = INF;
  for (long long int i = t1; i <= 10000; i++) {
    if (num[i] < count) {
      count = num[i];
      ans = t1 - vfilled[i];
    } else if (num[i] == count) {
      unsigned long long current = i;
      ans = min(ans, t1 - vfilled[i]);
    }
  }
  cout << count << " " << ans;
  return 0;
}
