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 dp[2][101][10001];
signed main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  memset(dp, -1, sizeof(dp));
  long long n, tot = 0;
  cin >> n;
  long long a[n], b[n], c[n];
  for (long long i = 0; i < n; i++) cin >> a[i], tot += a[i];
  for (long long i = 0; i < n; i++) cin >> b[i], c[i] = b[i];
  sort(c, c + n, greater<long long>());
  long long mx = -1;
  long long sum = 0;
  for (long long i = 0;; i++) {
    sum += c[i];
    if (sum >= tot) {
      mx = i + 1;
      cout << mx << ' ';
      break;
    }
  }
  dp[1][0][0] = 0;
  for (long long i = 0; i < n; i++) {
    long long cur = i & 1;
    long long prev = cur ^ 1;
    for (long long j = 0; j <= mx; j++) {
      for (long long k = 0; k <= 10000; k++) {
        dp[cur][j][k] = dp[prev][j][k];
        if (j && k >= b[i] && dp[prev][j - 1][k - b[i]] != -1)
          dp[cur][j][k] = max(dp[cur][j][k], dp[prev][j - 1][k - b[i]] + a[i]);
      }
    }
  }
  long long cur = ((n - 1) & 1);
  long long mx2 = 0;
  for (long long i = tot; i <= 10000; i++) mx2 = max(mx2, dp[cur][mx][i]);
  cout << tot - mx2;
}
