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;
template <typename T>
T getint() {
  T x = 0, p = 1;
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') p = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * p;
}
template <typename T1, typename T2>
bool umin(T1 &x, const T2 &y) {
  if (x > y) return x = y, true;
  return false;
}
template <typename T1, typename T2>
bool umax(T1 &x, const T2 &y) {
  if (x < y) return x = y, true;
  return false;
}
const int maxn = 1e6 + 10;
const int inf = (int)1e9 + 5;
const int mod = (int)1e9 + 7;
const long long llinf = (long long)1e18 + 5;
const long double pi = acos(-1.0);
int dp[2][101][101 * 101];
int main() {
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  int sum = 0;
  for (int i = (0); i < (n); i++) cin >> a[i], sum += a[i];
  vector<int> c(n);
  for (int i = (0); i < (n); i++) cin >> b[i], c[i] = b[i];
  sort((c).begin(), (c).end());
  reverse((c).begin(), (c).end());
  int x = sum;
  int k = 0;
  for (int i = (0); i < (c.size()); i++) {
    x -= c[i];
    k++;
    if (x <= 0) break;
  }
  for (int i = (0); i < (2); i++)
    for (int j = (0); j < (101); j++)
      for (int k = (0); k < (101 * 101); k++) dp[i][j][k] = -inf;
  dp[0][0][0] = 0;
  for (int i = (0); i < (n); i++) {
    for (int j = (0); j < (n + 1); j++) {
      for (int sum = (0); sum < (101 * 101); sum++) {
        if (sum >= b[i] && j > 0)
          umax(dp[1][j][sum], dp[0][j - 1][sum - b[i]] + a[i]);
        umax(dp[1][j][sum], dp[0][j][sum]);
      }
    }
    for (int j = (0); j < (n + 1); j++)
      for (int sum = (0); sum < (101 * 101); sum++) {
        dp[0][j][sum] = dp[1][j][sum];
        dp[1][j][sum] = -inf;
      }
  }
  int res = -inf;
  for (int j = (sum); j < (101 * 101); j++) {
    umax(res, dp[0][k][j]);
  }
  cout << k << ' ' << sum - res << endl;
  return 0;
}
