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 read() {
  int x = 0, o = 1;
  char ch = getchar();
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
const int N = 105;
int sum1, sum2, ans1, ans2, f[N][N * N];
struct ppx {
  int res, val;
} a[N];
inline bool cmp(ppx x, ppx y) {
  return x.val == y.val ? x.res > y.res : x.val > y.val;
}
int main() {
  int n = read();
  for (int i = 1; i <= n; ++i) {
    a[i].res = read();
    sum1 += a[i].res;
  }
  for (int i = 1; i <= n; ++i) {
    a[i].val = read();
    sum2 += a[i].val;
  }
  int now = 0;
  sort(a + 1, a + n + 1, cmp);
  for (int i = 1; i <= n; ++i) {
    now += a[i].val;
    if (now >= sum1) {
      ans1 = i;
      break;
    }
  }
  memset(f, -0x3f, sizeof(f));
  f[0][0] = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = sum2; j >= a[i].val; --j)
      for (int k = 1; k <= ans1; ++k)
        f[k][j] = max(f[k][j], f[k - 1][j - a[i].val] + a[i].res);
  for (int i = sum1; i <= sum2; ++i) ans2 = max(ans2, f[ans1][i]);
  printf("%d %d\n", ans1, sum1 - ans2);
  return 0;
}
