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;
namespace Sweet {
template <typename T>
inline void read(T &x) {
  char ch;
  int f = 1;
  while (ch = getchar(), ch > '9' || ch < '0') (ch == '-') && (f = -1);
  x = (ch ^ 48);
  while (ch = getchar(), ch >= '0' && ch <= '9') x = x * 10 + (ch ^ 48);
  x *= f;
}
template <typename T>
inline void write(T x) {
  static int stk[100], top = 0;
  if (x == 0) return (void)putchar('0');
  if (x < 0) x = -x, putchar('-');
  while (x) stk[++top] = x % 10, x /= 10;
  while (top) putchar(stk[top--] + '0');
}
const int N = 4e2 + 10;
struct E {
  int a, b;
  friend bool operator<(const E &a, const E &b) {
    return a.b == b.b ? a.a > b.a : a.b > b.b;
  }
} e[N];
inline void max(int &x, const int &y) { (x < y) && (x = y); }
int n, q, sigmaA, sigmaB, pre[N], f[N * N][N];
inline void main() {
  read(n);
  for (int i = 1; i <= n; ++i) read(e[i].a), sigmaA += e[i].a;
  for (int i = 1; i <= n; ++i) read(e[i].b), sigmaB += e[i].b;
  sort(e + 1, e + n + 1);
  for (int i = 1; i <= n; ++i) pre[i] = pre[i - 1] + e[i].b;
  memset(f, 0xcf, sizeof(f));
  f[0][0] = 0;
  int x = lower_bound(pre + 1, pre + n + 1, sigmaA) - pre;
  write(x), putchar(' ');
  for (int i = 1; i <= n; ++i) {
    for (int j = sigmaB; j >= e[i].b; --j) {
      for (int k = 1; k <= x; ++k) {
        max(f[j][k], f[j - e[i].b][k - 1] + e[i].a);
      }
    }
  }
  int ans = 0;
  for (int i = sigmaA; i <= sigmaB; ++i) max(ans, f[i][x]);
  write(sigmaA - ans);
}
}  // namespace Sweet
int main() {
  Sweet::main();
  return 0;
}
