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 = 100 + 10;
const int M = 100 * 100 + 10;
int n, a[N], b[N];
int dp[M], maxw[M];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  int sum = 0, total = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    sum += a[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> b[i];
    total += b[i];
  }
  memset(dp, -1, sizeof dp);
  memset(maxw, 0, sizeof maxw);
  dp[0] = maxw[0] = 0;
  for (int i = 0; i < n; ++i) {
    for (int k = total; k >= b[i]; --k) {
      if (dp[k - b[i]] != -1) {
        if (dp[k] == -1) {
          dp[k] = dp[k - b[i]] + 1;
          maxw[k] = maxw[k - b[i]] + a[i];
        } else {
          if (dp[k] > dp[k - b[i]] + 1) {
            dp[k] = dp[k - b[i]] + 1;
            maxw[k] = maxw[k - b[i]] + a[i];
          } else if (dp[k] == dp[k - b[i]] + 1) {
            maxw[k] = max(maxw[k], maxw[k - b[i]] + a[i]);
          }
        }
      }
    }
  }
  int num = 101, maxW = 0;
  for (int k = sum; k <= total; ++k) {
    if (dp[k] != -1) num = min(num, dp[k]);
  }
  for (int k = sum; k <= total; ++k) {
    if (dp[k] == num) maxW = max(maxW, maxw[k]);
  }
  cout << num << " " << sum - maxW << endl;
  return 0;
}
