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;
int n, sum = 0, f[30005], maxl, g[30005][105];
struct node {
  int a, b;
} t[105];
bool cmp(node x, node y) { return x.a > y.b; }
int main() {
  memset(f, 0x3f, sizeof(f));
  cin >> n;
  f[0] = 0;
  for (int i = 1; i <= n; i++) cin >> t[i].a, sum += t[i].a;
  for (int i = 1; i <= n; i++) cin >> t[i].b;
  for (int i = 1; i <= n; i++) {
    for (int j = sum; j >= 0; j--) {
      if (j >= t[i].b)
        f[j] = min(f[j], f[j - t[i].b] + 1);
      else
        f[j] = min(f[j], 1);
    }
  }
  cout << f[sum] << " ";
  memset(g, -0x3f, sizeof(g));
  g[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = sum + 10005; j >= t[i].b; j--) {
      for (int k = f[sum]; k >= 1; k--) {
        g[j][k] = max(g[j][k], g[j - t[i].b][k - 1] + t[i].a);
        if (j >= sum && k == f[sum]) maxl = max(maxl, g[j][k]);
      }
    }
  }
  cout << sum - maxl << endl;
}
