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 s[100], dp[101][10001];
int main() {
  int n;
  scanf("%d", &n);
  int ts = 0;
  for (int i = 0; i < (int)(n); ++i) scanf("%d", s + i), ts += s[i];
  for (int i = 0; i < (int)(n + 1); ++i)
    for (int j = 0; j < (int)(10001); ++j) dp[i][j] = -1e9;
  dp[0][0] = 0;
  for (int i = 0; i < (int)(n); ++i) {
    int sz;
    scanf("%d", &sz);
    for (int j = i; j >= 0; --j)
      for (int k = sz; k < (int)(10001); ++k)
        dp[j + 1][k] = max(dp[j + 1][k], dp[j][k - sz] + s[i]);
  }
  for (int i = 1; i < (int)(n + 1); ++i) {
    int an = -1;
    for (int j = ts; j < (int)(10001); ++j) an = max(an, dp[i][j]);
    if (an >= 0) {
      printf("%d %d\n", i, ts - an);
      return 0;
    }
  }
}
