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>
char BB[1 << 15], *SS = BB, *TT = BB;
inline int Read() {
  register int x = 0;
  register char ch =
      (SS == TT && (TT = (SS = BB) + fread(BB, 1, 1 << 15, stdin), TT == SS)
           ? EOF
           : *SS++);
  while (ch < '0' || ch > '9')
    ch = (SS == TT && (TT = (SS = BB) + fread(BB, 1, 1 << 15, stdin), TT == SS)
              ? EOF
              : *SS++);
  while (ch <= '9' && ch >= '0') {
    x = x * 10 + (ch ^ 48);
    ch = (SS == TT && (TT = (SS = BB) + fread(BB, 1, 1 << 15, stdin), TT == SS)
              ? EOF
              : *SS++);
  }
  return x;
}
inline int Min(const int x, const int y) { return x < y ? x : y; }
inline int ABS(const int x) { return x < 0 ? ~x + 1 : 0x3f3f3f3f3f; }
int a[100 + 5], b[100 + 5], n, m, f[10000 + 42], g[10000 + 42], ans;
int main() {
  n = Read();
  for (register int i = 1; i <= n; ++i) m += (a[i] = Read());
  for (register int i = 1; i <= n; ++i) b[i] = Min(Read(), m);
  for (register int i = m + 100; i >= 1; --i) f[i] = n;
  for (register int i = 1; i <= n; ++i)
    for (register int j = m + 100; j >= b[i]; --j)
      if (f[j - b[i]] + 1 < f[j]) {
        if (j <= m) {
          f[j] = f[j - b[i]] + 1;
          g[j] = g[j - b[i]] + b[i] - a[i];
        } else {
          f[j] = f[j - b[i]] + 1;
          g[j] = g[j - b[i]] + b[i] - a[i] + m - j;
        }
      } else if (f[j - b[i]] + 1 == f[j]) {
        if (j <= m)
          g[j] = Min(g[j], g[j - b[i]] + b[i] - a[i]);
        else
          g[j] = Min(g[j], g[j - b[i]] + b[i] - a[i] + m - j);
      }
  f[0] = n;
  for (register int i = m + 100; i >= m; --i)
    if (f[ans] > f[i])
      ans = i;
    else if (f[ans] == f[i] && g[ans] > g[i])
      ans = i;
  printf("%d %d", f[ans], g[ans]);
  return 0;
}
