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;
pair<int, int> a[105];
int dp[3][103][10005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int tv = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].second;
    tv += a[i].second;
  }
  for (int i = 1; i <= n; i++) cin >> a[i].first;
  sort(a + 1, a + n + 1);
  int tv1 = 0;
  int k;
  for (int i = n; i >= 1; i--) {
    tv1 += a[i].first;
    if (tv1 >= tv) {
      k = n - i + 1;
      break;
    }
  }
  int ans = 0;
  for (int j = 0; j <= k; j++) {
    for (int u = 0; u <= 10000; u++) {
      dp[2][j][u] = -100000000;
    }
  }
  dp[2][0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      for (int u = 0; u <= 10000; u++) {
        if (u >= a[i].first) {
          dp[1][j][u] =
              max(dp[2][j][u], dp[2][j - 1][u - a[i].first] + a[i].second);
        } else
          dp[1][j][u] = dp[2][j][u];
        if (j == k && u >= tv) ans = max(ans, dp[1][j][u]);
      }
    }
    for (int j = 1; j <= k; j++) {
      for (int u = 1; u <= 10000; u++) dp[2][j][u] = dp[1][j][u];
    }
  }
  cout << k << '\n';
  cout << tv - ans << '\n';
}
