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;
constexpr int INF = 1000000000;
int N;
pair<int, int> bottles[100];
map<pair<int, int>, pair<int, int> > memo;
template <typename T, typename U>
pair<T, U> operator+(const pair<T, U>& a, const pair<T, U>& b) {
  return {a.first + b.first, a.second + b.second};
}
pair<int, int> f(int p, int v) {
  if (v <= 0) return pair<int, int>(0, v);
  if (p >= N) return pair<int, int>(INF, INF);
  if (memo.find({p, v}) != memo.end()) return memo[{p, v}];
  pair<int, int> r1 = f(p + 1, v - bottles[p].second) +
                      pair<int, int>(1, bottles[p].second - bottles[p].first);
  pair<int, int> r2 = f(p + 1, v);
  return memo[{p, v}] = min(r1, r2);
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", &bottles[i].first);
  for (int i = 0; i < N; i++) scanf("%d", &bottles[i].second);
  int vol = 0;
  for (int i = 0; i < N; i++) vol += bottles[i].first;
  pair<int, int> res = f(0, vol);
  printf("%d %d\n", res.first, res.second);
}
