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;
const long long MAXN = 111;
long long a[MAXN], b[MAXN];
pair<long long, long long> dp[MAXN][MAXN * MAXN];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, s = 0;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i], s += a[i];
  for (long long i = 0; i < n; i++) cin >> b[i];
  dp[n - 1][0] = {0, 0};
  for (long long i = 1; i <= b[n - 1]; i++) dp[n - 1][i] = {1, -a[n - 1]};
  for (long long i = b[n - 1] + 1; i <= s; i++) dp[n - 1][i] = {1e18, a[n - 1]};
  for (long long i = n - 2; i > -1; i--) {
    dp[i][0] = {0, 0};
    for (long long j = 1; j <= s; j++) {
      dp[i][j] = dp[i + 1][j];
      pair<long long, long long> p = dp[i + 1][max(0ll, j - b[i])];
      p.first++;
      p.second -= a[i];
      dp[i][j] = min(dp[i][j], p);
    }
  }
  cout << dp[0][s].first << ' ' << s + dp[0][s].second;
}
