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;
long long n, a[101], b[101];
vector<vector<long long>> dp(101, vector<long long>(10001, -10000000000));
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (long long i = 1; i <= n; i++) {
    cin >> b[i];
  }
  dp[0][0] = 0;
  vector<vector<long long>> dp_new(101, vector<long long>(10001));
  dp_new = dp;
  dp[1][b[1]] = a[1];
  for (long long i = 2; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      for (long long k = 1; k <= 10000; k++) {
        dp_new[j][k] = dp[j][k];
        if (k > b[i]) {
          dp_new[j][k] = max(dp_new[j][k], dp[j - 1][k - b[i]] + a[i]);
        }
      }
    }
    dp_new[1][b[i]] = max(dp_new[1][b[i]], a[i]);
    dp = dp_new;
  }
  sort(b + 1, b + n + 1);
  long long sm = 0;
  for (long long i = 1; i <= n; i++) sm += a[i];
  long long k;
  for (long long i = n; i >= 0; i--) {
    if (sm <= 0) {
      k = n - i;
      break;
    }
    sm -= b[i];
  }
  sm = 0;
  for (long long i = 1; i <= n; i++) sm += a[i];
  long long ans = 10000000000;
  for (long long i = sm; i <= 10000; i++) {
    ans = min(ans, sm - dp[k][i]);
  }
  cout << k << " " << ans << '\n';
}
