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>
struct lx {
  int x, y;
} a[105];
bool cmp(lx a, lx b) { return a.y > b.y; }
inline int min(int a, int b) {
  a -= b;
  return b + (a & (a >> 31));
}
int n, m, v, sum, ans, f[105][10005];
inline int read() {
  int c = getchar(), t = 0;
  for (; c < 48 || c > 57; c = getchar())
    ;
  do {
    t = (t << 1) + (t << 3) + c - 48;
    c = getchar();
  } while (48 <= c && c <= 57);
  return t;
}
int main() {
  n = read();
  memset(f, 127, sizeof f);
  for (int i = 1; i <= n; ++i) sum += a[i].x = read();
  for (int i = 1; i <= n; ++i) a[i].y = read();
  std::sort(a + 1, a + n + 1, cmp);
  ans = f[0][0] = sum;
  for (int i = 1; i <= n && v < sum; ++i) v += a[i].y, ++m;
  for (int i = 1; i <= n; ++i)
    for (int j = min(i, m); j; --j)
      for (int k = a[i].y; k <= v; ++k)
        f[j][k] = min(f[j][k], f[j - 1][k - a[i].y] - a[i].x);
  for (int i = sum; i <= v; ++i) ans = min(ans, f[m][i]);
  printf("%d %d", m, ans);
}
