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;
pair<int, int> dp[10005];
int a[105], b[105];
int main() {
  int n;
  scanf("%d", &n);
  int cap = 0, mx = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    cap += a[i];
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &b[i]);
    mx += b[i];
  }
  for (int i = 1; i <= mx; i++) dp[i] = make_pair(-1, -1);
  for (int i = 0; i < n; i++) {
    for (int j = mx; j >= 0; j--) {
      if (dp[j].first == -1) continue;
      int nxt = j + b[i];
      if (dp[nxt].first == -1)
        dp[nxt] = make_pair(dp[j].first + 1, dp[j].second + a[i]);
      else if (dp[nxt].first > dp[j].first + 1)
        dp[nxt] = make_pair(dp[j].first + 1, dp[j].second + a[i]);
      else if (dp[nxt].first == dp[j].first + 1)
        dp[nxt].second = max(dp[nxt].second, dp[j].second + a[i]);
    }
  }
  pair<int, int> mn = make_pair((int)1e9, 0);
  for (int i = cap; i <= mx; i++) {
    if (dp[i].first == -1) continue;
    if (dp[i].first < mn.first)
      mn = dp[i];
    else if (dp[i].first == mn.first)
      mn.second = max(mn.second, dp[i].second);
  }
  printf("%d %d\n", mn.first, cap - mn.second);
  return 0;
}
