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 dp[105][10010];
int a[105], b[105];
int main() {
  ios_base::sync_with_stdio(false);
  int i, j, k, n;
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < n; i++) cin >> b[i];
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  int s, t;
  s = t = 0;
  for (i = 0; i < n; i++) {
    for (j = i; j >= 0; j--) {
      for (k = 0; k <= t; k++) {
        if (dp[j][k] != -1) {
          dp[j + 1][k + b[i]] = max(dp[j + 1][k + b[i]], dp[j][k] + a[i]);
        }
      }
    }
    s += a[i];
    t += b[i];
  }
  sort(b, b + n);
  reverse(b, b + n);
  int ans1, ans2;
  ans1 = k = 0;
  ans2 = 1099999;
  while (k < s) {
    k += b[ans1];
    ans1++;
  }
  for (i = s; i <= t; ++i) ans2 = min(ans2, s - dp[ans1][i]);
  cout << ans1 << " " << ans2 << endl;
  return 0;
}
