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>
#pragma comment(linker, "/STACK:167772160000")
using namespace std;
pair<int, int> bottles[110];
int totAmount;
pair<int, int> dp[110][11000];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < (n); i++) {
    cin >> bottles[i].second;
  }
  for (int i = 0; i < (n); i++) cin >> bottles[i].first;
  for (int i = 0; i < (n + 1); i++)
    for (int j = 0; j < (11000); j++) dp[i][j] = pair<int, int>(1E8, 1E8);
  dp[0][0] = pair<int, int>(0, 0);
  for (int i = 0; i < (n); i++) {
    totAmount += bottles[i].second;
    for (int j = 0; j < (11000); j++) {
      dp[i + 1][j] = min(dp[i][j], dp[i + 1][j]);
      for (int k = 0; k <= bottles[i].first; k++) {
        if (k <= bottles[i].second)
          dp[i + 1][j + k] =
              min(dp[i + 1][j + k],
                  pair<int, int>(dp[i][j].first + 1, dp[i][j].second));
        else
          dp[i + 1][j + k] =
              min(dp[i + 1][j + k],
                  pair<int, int>(dp[i][j].first + 1,
                                 dp[i][j].second + k - bottles[i].second));
      }
    }
  }
  cout << dp[n][totAmount].first << " " << dp[n][totAmount].second << endl;
  return 0;
}
