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 inf = 2e9;
int mod = 1e9 + 7;
char en = '\n';
int arr[105], arr2[105], temp[105];
unordered_map<int, int> par[105][105];
int recur(int i, int k, int sum) {
  if (par[i][k].count(sum)) return par[i][k][sum];
  if (k == 0) {
    if (sum <= 0)
      return 0;
    else
      return -inf;
  }
  if (i == 0) {
    return -inf;
  }
  int val =
      max(recur(i - 1, k, sum), arr[i] + recur(i - 1, k - 1, sum - arr2[i]));
  par[i][k][sum] = val;
  return val;
}
int main() {
  int n;
  cin >> n;
  int sum1 = 0;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    sum1 += arr[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> arr2[i];
    temp[i] = arr2[i];
  }
  int k;
  int curr = 0;
  sort(temp + 1, temp + n + 1, greater<int>());
  for (int i = 1; i <= n; i++) {
    curr += temp[i];
    if (curr >= sum1) {
      k = i;
      break;
    }
  }
  int t = recur(n, k, sum1);
  cout << k << " " << sum1 - t << en;
  return 0;
}
