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 = 1e6;
inline int getint() {
  int w = 0, q = 0;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') q = 1, c = getchar();
  while (c >= '0' && c <= '9') w = w * 10 + c - '0', c = getchar();
  return q ? -w : w;
}
int a[110];
int b[110];
int n;
int dp[11000][2];
int main() {
  n = getint();
  int sum = 0;
  for (int i = 0; i < n; i++) {
    a[i] = getint();
    sum += a[i];
  }
  for (int i = 0; i < n; i++) {
    b[i] = getint();
  }
  for (int i = 0; i <= sum; i++) {
    dp[i][0] = 0X3F7F7F7F;
    dp[i][1] = 0;
  }
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = sum; j > 0; j--) {
      int t = max(j - b[i], 0);
      if (dp[j][0] > dp[t][0] + 1) {
        dp[j][0] = dp[t][0] + 1;
        dp[j][1] = dp[t][1] + a[i];
      } else if (dp[j][0] == dp[t][0] + 1 && dp[j][1] < dp[t][1] + a[i]) {
        dp[j][1] = dp[t][1] + a[i];
      }
    }
  }
  printf("%d %d\n", dp[sum][0], sum - dp[sum][1]);
  return 0;
}
