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;
inline int ra() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int a[100005], b[100005];
int f[100005], size[100005];
int main() {
  int n = ra(), sum = 0, tot = 0;
  memset(size, 0x7, sizeof(size));
  size[0] = 0;
  for (int i = 1; i <= n; i++) a[i] = ra(), sum += a[i];
  for (int i = 1; i <= n; i++) b[i] = ra(), tot += b[i];
  for (int i = 1; i <= n; i++)
    for (int j = tot; j >= 0; j--) {
      if (size[j] + 1 == size[j + b[i]])
        f[j + b[i]] = max(f[j + b[i]], f[j] + a[i]);
      if (size[j] + 1 < size[j + b[i]]) {
        size[j + b[i]] = size[j] + 1;
        f[j + b[i]] = f[j] + a[i];
      }
    }
  int anssz = 0x3f3f3f3f, ans;
  for (int i = sum; i <= tot; i++) {
    if (anssz == size[i]) ans = max(ans, f[i]);
    if (anssz > size[i]) {
      anssz = size[i];
      ans = f[i];
    }
  }
  printf("%d %d", anssz, sum - ans);
  return 0;
}
