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[101][101 * 101];
int data[101];
int s[101];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  int sum = 0;
  vector<pair<int, int> > v;
  for (int i = 1; i <= n; i++) cin >> data[i], sum += data[i];
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    v.push_back(pair<int, int>(a, data[i]));
  }
  sort(v.begin(), v.end(), greater<pair<int, int> >());
  for (int i = 1; i <= n; i++) s[i] = s[i - 1] + v[i - 1].first;
  int idx = lower_bound(s + 1, s + n + 1, sum) - s;
  memset(dp, 192, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j > 0; j--) {
      for (int k = v[i - 1].second; k <= sum; k++)
        dp[j][k] =
            max(dp[j][k], dp[j - 1][k - v[i - 1].second] + v[i - 1].first);
    }
  }
  cout << idx << ' ';
  for (int i = sum; i >= 0; i--) {
    if (dp[idx][i] >= sum) {
      cout << sum - i << "\n";
      break;
    }
  }
  return 0;
}
