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;
struct ura {
  int a, b;
} v[105];
int d[10005][105];
bool cmp(ura a, ura b) { return a.b < b.b; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long i, j = 0, n, mn = 9999999999LL, k = 0, pz = -1, s1 = 0, s2 = 0,
               sc = 0;
  cin >> n;
  if (n == 1) {
    cout << 1 << ' ' << 0;
    return 0;
  }
  for (i = 1; i <= n; ++i) {
    cin >> v[i].a;
    s1 += v[i].a;
  }
  for (i = 1; i <= n; ++i) {
    cin >> v[i].b;
    s2 += v[i].b;
  }
  sort(v + 1, v + n + 1, cmp);
  for (i = n; i >= 1; --i) {
    sc += v[i].b;
    ++k;
    if (sc >= s1) break;
  }
  cout << k << ' ';
  k = n - k;
  for (pz = 1; pz <= n; ++pz) {
    for (j = k - 1; j >= 1; --j) {
      for (i = 1; i <= s2 - s1 - v[pz].b; ++i) {
        if (d[i][j]) {
          if (d[i + v[pz].b][j + 1])
            d[i + v[pz].b][j + 1] =
                min(d[i + v[pz].b][j + 1], d[i][j] + v[pz].a);
          else
            d[i + v[pz].b][j + 1] = d[i][j] + v[pz].a;
        }
      }
    }
    if (d[v[pz].b][1])
      d[v[pz].b][1] = min(d[v[pz].b][1], v[pz].a);
    else
      d[v[pz].b][1] = v[pz].a;
  }
  for (i = 1; i <= s2 - s1; ++i) {
    if (d[i][k])
      if (d[i][k] < mn) mn = d[i][k];
  }
  if (mn == 9999999999LL)
    cout << 0;
  else
    cout << mn;
}
