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 n;
int a[105], b[105];
pair<int, int> dp[105][105 * 105];
bool seen[105][105 * 105];
int tsa;
pair<int, int> f(int i, int sb) {
  if (seen[i][sb]) return dp[i][sb];
  seen[i][sb] = true;
  if (i == n) {
    if (sb >= tsa) {
      return dp[i][sb] = make_pair(0, 0);
    } else {
      return dp[i][sb] = make_pair(105, 0);
    }
  } else {
    auto p = f(i + 1, sb);
    p.second += a[i];
    auto op = f(i + 1, sb + b[i]);
    op.first += 1;
    return dp[i][sb] = min(op, p);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); i++) {
    scanf("%d", a + i);
    tsa += a[i];
  }
  for (int i = (0); i < (n); i++) scanf("%d", b + i);
  auto p = f(0, 0);
  printf("%d %d\n", p.first, p.second);
  return 0;
}
