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 = 200010;
inline int in() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long lin() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
struct Info {
  int a, b;
} inp[205];
int tot;
int n;
pair<int, int> dp[105][10005];
pair<int, int> go(int now, int rem) {
  if (rem <= 0) return make_pair(0, 0);
  if (now > n) return make_pair(100000, 100000);
  if (dp[now][rem] != make_pair(-1, -1)) return dp[now][rem];
  pair<int, int> a;
  a = go(now + 1, rem - inp[now].b);
  pair<int, int> p, q;
  p.first = 1 + a.first;
  if (rem - inp[now].b >= 0)
    p.second = (inp[now].b - inp[now].a) + a.second;
  else
    p.second = (rem - inp[now].a) + a.second;
  q = go(now + 1, rem);
  if (q.first < p.first)
    p = q;
  else if (q.first == p.first) {
    if (q.second < p.second) p = q;
  }
  return dp[now][rem] = p;
}
int main() {
  n = in();
  for (int i = 1; i <= n; i++) {
    inp[i].a = in();
    tot += inp[i].a;
  }
  for (int i = 1; i <= n; i++) {
    inp[i].b = in();
  }
  memset(dp, -1, sizeof dp);
  pair<int, int> ans = go(1, tot);
  printf("%d %d\n", ans.first, ans.second);
}
