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;
pair<int, int> dp[20001];
pair<int, int> dp_copy[20001];
int a[100];
int b[100];
void copy_dp() {
  for (int i = -10000; i <= 10000; i++) {
    dp_copy[i + 10000] = dp[i + 10000];
    dp[i + 10000] = make_pair(1e9, 1e9);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int t;
  for (int i = 0; i < n; i++) scanf("%d", &t), a[i] = t;
  for (int i = 0; i < n; i++) scanf("%d", &t), b[i] = t;
  copy_dp();
  dp[0 + 10000] = make_pair(0, 0);
  for (int i = 0; i < n; i++) {
    copy_dp();
    for (int j = -10000 + a[i]; j <= 10000; j++)
      dp[j - a[i] + 10000] = min(dp[j - a[i] + 10000],
                                 make_pair(dp_copy[j + 10000].first,
                                           dp_copy[j + 10000].second + a[i]));
    for (int j = -10000; j <= 10000; j++)
      dp[j + b[i] - a[i] + 10000] = min(
          dp[j + b[i] - a[i] + 10000],
          make_pair(dp_copy[j + 10000].first + 1, dp_copy[j + 10000].second));
  }
  pair<int, int> res_min = make_pair(1e9, 1e9);
  for (int i = 0; i <= 10000; i++) res_min = min(dp[i + 10000], res_min);
  printf("%d %d\n", res_min.first, res_min.second);
  return 0;
}
