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;
int a[105], b[105], sum;
int f[105][105 * 105], g[105][105 * 105];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  memset(f, 0x3f, sizeof(f));
  f[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    sum += a[i];
    for (int j = 0; j < 105 * 105; j++) {
      f[i][j] = f[i - 1][j];
      g[i][j] = g[i - 1][j];
      if (j - b[i] >= 0) {
        if (f[i][j] > f[i - 1][j - b[i]] + 1) {
          f[i][j] = f[i - 1][j - b[i]] + 1;
          g[i][j] = g[i - 1][j - b[i]] + a[i];
        } else if (f[i][j] == f[i - 1][j - b[i]] + 1) {
          g[i][j] = max(g[i][j], g[i - 1][j - b[i]] + a[i]);
        }
      }
    }
  }
  int x = 1e9, y = 0;
  for (int i = sum; i < 105 * 105; i++) {
    if (f[n][i] < x) {
      x = f[n][i], y = g[n][i];
    } else if (f[n][i] == x) {
      y = max(y, g[n][i]);
    }
  }
  printf("%d %d", x, sum - y);
}
