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 = 120;
int w[maxn], v[maxn];
pair<int, int> dp[maxn * maxn];
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    int sum = 0, need = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &v[i]);
      need += v[i];
    }
    for (int i = 0; i < n; i++) {
      scanf("%d", &w[i]);
      sum += w[i];
    }
    for (int i = 0; i <= sum; i++) dp[i] = make_pair(1e9, 0);
    dp[0] = make_pair(0, 0);
    for (int i = 0; i < n; i++) {
      for (int j = sum; j >= w[i]; j--) {
        pair<int, int> tmp = dp[j - w[i]];
        if (dp[j].first > dp[j - w[i]].first + 1) {
          dp[j].first = dp[j - w[i]].first + 1;
          dp[j].second = dp[j - w[i]].second + v[i];
        } else if (dp[j].first == dp[j - w[i]].first + 1) {
          dp[j].second = max(dp[j].second, dp[j - w[i]].second + v[i]);
        }
      }
    }
    pair<int, int> ans;
    ans = make_pair(n, 0);
    for (int i = need; i <= sum; i++) {
      if (dp[i].first < ans.first)
        ans = dp[i];
      else if (dp[i].first == ans.first) {
        ans.second = max(ans.second, dp[i].second);
      }
    }
    printf("%d %d\n", ans.first, need - ans.second);
  }
  return 0;
}
