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;
int N, M, ansx, ansy, Max, now, pre;
int f[2][110][10010];
class rec {
 public:
  int v, r;
} P[110];
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &P[i].r);
  for (int i = 1; i <= N; i++) scanf("%d", &P[i].v);
  Max = 0;
  pre = 1;
  now = 0;
  memset(f[now], 230, sizeof(f[now]));
  f[now][0][0] = 0;
  for (int i = 1; i <= N; i++) {
    swap(now, pre);
    memset(f[now], 230, sizeof(f[now]));
    for (int j = 0; j < i; j++)
      for (int k = 0; k <= Max; k++)
        if (f[pre][j][k] > -1e8) {
          f[now][j][k + P[i].r] =
              max(f[now][j][k + P[i].r], f[pre][j][k] - P[i].r);
          f[now][j + 1][k] =
              max(f[now][j + 1][k], f[pre][j][k] + P[i].v - P[i].r);
        }
    Max += P[i].v;
  }
  ansx = -1;
  for (int j = 0; j <= N && ansx == -1; j++)
    for (int k = 0; k <= Max; k++)
      if (f[now][j][k] >= 0) {
        ansx = j;
        ansy = k;
        break;
      }
  printf("%d %d\n", ansx, ansy);
  return 0;
}
