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 MAXN = 100;
int memo[MAXN + 1][100 * MAXN + 1];
bool vis[MAXN + 1][100 * MAXN + 1];
int main() {
  int n;
  cin >> n;
  int total = 0;
  int arr[n];
  int kek[n];
  for (int i = 0; i != n; ++i) {
    cin >> arr[i];
    total += arr[i];
  }
  for (int i = 0; i != n; ++i) cin >> kek[i];
  vis[0][0] = 1;
  for (int i = 0; i != n; ++i) {
    for (int j = n; j != 0; --j) {
      for (int k = 100 * n; k >= kek[i]; --k) {
        if (vis[j - 1][k - kek[i]]) {
          vis[j][k] = 1;
          memo[j][k] = max(memo[j][k], memo[j - 1][k - kek[i]] + arr[i]);
        }
      }
    }
  }
  int ans1, ans2 = 1000000000;
  for (int i = 1; i <= n; ++i) {
    bool done = 0;
    for (int j = total; j <= 100 * n; ++j) {
      if (vis[i][j]) {
        done = 1;
        ans1 = i;
        ans2 = min(ans2, total - memo[i][j]);
      }
    }
    if (done) break;
  }
  cout << ans1 << ' ' << ans2 << '\n';
  return 0;
}
