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;
const int N = 105;
const int M = 10005;
int n, m, k, a[N], b[N];
int sa, sb, id[N];
int f[N][M], ans;
bool cmp(int i, int j) { return b[i] > b[j]; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) id[i] = i;
  sa = sb = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    sa += a[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    sb += b[i];
  };
  m = sb - sa;
  sort(id + 1, id + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    sa -= b[id[i]];
    if (sa <= 0) {
      k = i;
      break;
    }
  }
  k = n - k;
  memset(f, -1, sizeof(f));
  f[0][0] = 0;
  for (int h = 1; h <= n; h++)
    for (int i = min(k, h); i >= 1; i--)
      for (int j = m - b[h]; j >= 0; j--)
        if (f[i - 1][j] != -1 &&
            (f[i][j + b[h]] == -1 || f[i - 1][j] + a[h] < f[i][j + b[h]]))
          f[i][j + b[h]] = f[i - 1][j] + a[h];
  ans = M;
  for (int i = 0; i <= m; i++)
    if (f[k][i] != -1) ans = min(ans, f[k][i]);
  printf("%d %d\n", n - k, ans);
  return 0;
}
