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 k, n, m, sum, cur;
pair<int, int> a[110];
int f[110][10010], mn[110][10010];
bool Cmp(pair<int, int> a, pair<int, int> b) {
  return a.second > b.second || (a.second == b.second && a.first > b.first);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].first), sum += a[i].first;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].second);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < i; j++)
      for (int k = 0; k < sum; k++)
        if (!k || mn[j][k]) {
          int t = mn[j][k] + 1;
          if (!mn[i][k + a[i].second] || t < mn[i][k + a[i].second] ||
              (t == mn[i][k + a[i].second] &&
               f[j][k] + a[i].first > f[i][k + a[i].second]))
            mn[i][k + a[i].second] = t,
                      f[i][k + a[i].second] = f[j][k] + a[i].first;
        }
  }
  int Mn = n + 1, Mx = 0;
  for (int i = 1; i <= n; i++)
    for (int j = sum; j <= 10000; j++)
      if (mn[i][j] && mn[i][j] < Mn || (mn[i][j] == Mn && f[i][j] > Mx))
        Mn = mn[i][j], Mx = f[i][j];
  printf("%d %d\n", Mn, sum - Mx);
}
