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;
int n, a[110], b[110], sum = 0;
pair<int, int> dp[110][11000], q = {-1, -1};
pair<int, int> Rec(int i, int W) {
  if (i == n) {
    if (W >= sum)
      return {0, 0};
    else
      return {1e9, 1e9};
  }
  if (dp[i][W] != q) return dp[i][W];
  pair<int, int> Res1 = {1e9, 1e9}, Res2 = {1e9, 1e9};
  Res1 = {0 + Rec(i + 1, W).first, a[i] + Rec(i + 1, W).second};
  Res2 = {1 + Rec(i + 1, W + b[i]).first, 0 + Rec(i + 1, W + b[i]).second};
  return dp[i][W] = min(Res1, Res2);
}
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], sum += a[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  cout << Rec(0, 0).first << " " << Rec(0, 0).second;
  return 0;
}
