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, typename U>
inline void chkmax(T& x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
inline void chkmin(T& x, U y) {
  if (y < x) x = y;
}
int k, t, a[111], tot, id[111], dp[111][111 * 111], b[111], s, n;
bool cmp(int u, int v) { return b[u] > b[v]; }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), tot += a[i];
  for (int i = 0; i < n; i++) scanf("%d", &b[i]), id[i] = i;
  sort(id, id + n, cmp);
  for (int i = 0; i < n; i++) {
    s += b[id[i]];
    if (s >= tot) {
      k = i + 1;
      break;
    }
  }
  for (int i = 0; i <= k; i++)
    for (int j = 0; j < 111 * 111; j++) dp[i][j] = -1e8;
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = k; j >= 1; j--)
      for (int l = 0; l < 111 * 111 - b[i]; l++) {
        chkmax(dp[j][l + b[i]], dp[j - 1][l] + a[i]);
      }
  }
  t = 1e8;
  for (int j = tot; j < 111 * 111; j++) chkmin(t, tot - dp[k][j]);
  cout << k << " " << t << endl;
  return 0;
}
