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 int N = 1e4 + 4;
pair<int, int> DP[101][2 * N];
int F[101][101];
int a[101], b[101];
const int inf = 1e5 + 5;
pair<int, int>& dp(int i, int j) {
  if (j < 0) return DP[i][N + abs(j)];
  return DP[i][j];
}
int n;
pair<int, int> rec(int idx, int volume_free) {
  if (dp(idx, volume_free).first != -1) return dp(idx, volume_free);
  if (idx == n) {
    if (volume_free >= 0) return make_pair(0, 0);
    return {inf, 0};
  }
  pair<int, int> res;
  pair<int, int> ff = rec(idx + 1, volume_free + b[idx] - a[idx]);
  res.first = ff.first + 1;
  res.second = ff.second;
  ff = rec(idx + 1, volume_free - a[idx]);
  res = min(res, {ff.first, ff.second + a[idx]});
  return dp(idx, volume_free) = res;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i) cin >> b[i];
  for (int i = 0; i < 101; ++i)
    for (int j = -N; j < N; ++j) dp(i, j).first = -1;
  cout << rec(0, 0).first << ' ' << rec(0, 0).second << '\n';
  return 0;
}
