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;
bool comp(const pair<int, int>& a, const pair<int, int>& b) {
  if (a.second != b.second) {
    return a.second > b.second;
  }
  return false;
}
const int inf = 1e9;
int main() {
  int n;
  cin >> n;
  vector<pair<int, int> > v(n);
  int x = 0;
  for (int i = 0; i < n; ++i) {
    cin >> v[i].first;
    x += v[i].first;
  }
  for (int i = 0; i < n; ++i) {
    cin >> v[i].second;
  }
  sort((v).begin(), (v).end(), comp);
  int k = 0;
  while (x > 0) {
    x -= v[k].second;
    ++k;
  }
  x = 0;
  for (int i = 0; i < n; ++i) {
    x += v[i].first;
  }
  cout << k << " ";
  vector<vector<int> > dp(x + 1, vector<int>(k + 1, -inf));
  dp[0][0] = 0;
  for (int i = 0; i < n; ++i) {
    vector<vector<int> > dp_new = dp;
    for (int sum = 0; sum <= x; ++sum) {
      for (int taken = 0; taken <= k; ++taken) {
        int new_sum = min(x, sum + v[i].second);
        if (taken + 1 <= k) {
          dp_new[new_sum][taken + 1] =
              max(dp_new[new_sum][taken + 1], dp[sum][taken] + v[i].first);
        }
      }
    }
    dp.swap(dp_new);
  }
  cout << x - dp[x][k] << "\n";
}
