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;
template <typename T>
inline void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  x *= f;
}
const int maxn = 110;
int n;
int tot, res, ans;
struct bottle {
  int a, b;
  friend bool operator<(bottle a, bottle b) { return a.b > b.b; }
} bt[maxn];
int f[maxn][maxn * maxn], mx;
inline void dp() {
  memset(f, -100, sizeof(f));
  f[0][0] = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = mx; j >= bt[i].b; --j)
      for (int k = 1; k <= res; ++k)
        f[k][j] = max(f[k][j], f[k - 1][j - bt[i].b] + bt[i].a);
}
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) {
    read(bt[i].a);
    tot += bt[i].a;
  }
  for (int i = 1; i <= n; ++i) {
    read(bt[i].b);
    mx += bt[i].b;
  }
  sort(bt + 1, bt + n + 1);
  ans = tot;
  for (int i = 1; i <= n; ++i) {
    if (tot > bt[i].b)
      tot -= bt[i].b;
    else {
      res = i;
      break;
    }
  }
  dp();
  int maxx = 0;
  for (int i = ans; i <= mx; ++i) maxx = max(maxx, f[res][i]);
  cout << res << ' ' << ans - maxx;
  return 0;
}
