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 = 105;
int n, a[N], b[N];
pair<int, int> dp[N][N * N];
void init() {
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N * N; j++) dp[i][j] = {-1, -1};
  }
}
pair<int, int> sum(pair<int, int> p1, pair<int, int> p2) {
  return make_pair(p1.first + p2.first, p1.second + p2.second);
}
pair<int, int> solve(int idx, int rem) {
  if (rem <= 0) return {0, 0};
  if (idx == n) return {(int)1 << 30, (int)1 << 30};
  if (dp[idx][rem] != make_pair(-1, -1)) return dp[idx][rem];
  int mn = min(b[idx] - a[idx], rem - a[idx]);
  return dp[idx][rem] =
             min(sum(solve(idx + 1, rem - mn - a[idx]), make_pair(1, mn)),
                 solve(idx + 1, rem));
}
int main() {
  scanf("%d", &n);
  int tot = 0;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), tot += a[i];
  for (int i = 0; i < n; i++) scanf("%d", &b[i]);
  init();
  pair<int, int> ans = solve(0, tot);
  printf("%d %d", ans.first, ans.second);
}
