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 MAX = 1e4 + 10;
struct Bottle {
  int v, n;
} a[110];
int DP[110][MAX];
int main() {
  int n, sum_n = 0, sum_v = 0, maxx = 0, num = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i].n);
    sum_n += a[i].n;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i].v);
    sum_v += a[i].v;
  }
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j; j--)
      for (int k = sum_v; k >= a[i].v; k--)
        if (DP[j - 1][k - a[i].v] || k == a[i].v)
          DP[j][k] = max(DP[j][k], DP[j - 1][k - a[i].v] + a[i].n);
  for (int i = 1; i <= n; i++) {
    for (int j = sum_n; j <= sum_v; j++)
      if (DP[i][j] > maxx) {
        maxx = DP[i][j];
        num = i;
      }
    if (maxx != 0) break;
  }
  printf("%d %d\n", num, sum_n - maxx);
  return 0;
}
