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 a[101], b[101], dp[101][101 * 101], inf = 1E9;
int main() {
  int n, m, i, j, k;
  pair<int, int> ans = make_pair(inf, inf);
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (i = 1; i <= n; i++) scanf("%d", &b[i]);
  for (i = 0; i < 101; i++)
    for (j = 0; j < 101 * 101; j++) dp[i][j] = -inf;
  dp[0][0] = 0;
  for (k = 1, m = 0; k <= n; k++) {
    b[k] -= a[k];
    m += a[k];
    for (i = k; i > 0; i--)
      for (j = m; j >= a[k]; j--)
        dp[i][j] = max(dp[i][j], dp[i - 1][j - a[k]] + b[k]);
  }
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++)
      if (j + dp[i][j] >= m) ans = min(ans, make_pair(i, m - j));
  printf("%d %d\n", ans.first, ans.second);
  return 0;
}
