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;
inline int max(int x, int y) { return x > y ? x : y; }
int n;
struct node {
  int x, y;
  bool operator<(const node &rhs) const { return y > rhs.y; }
} p[250];
int d[250][250 * 250];
int main() {
  int sum = 0, aans = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    cin >> p[i].x;
    sum += p[i].x;
  }
  for (int i = 1; i <= n; i++) {
    cin >> p[i].y;
    aans += p[i].y;
  }
  sort(p + 1, p + 1 + n);
  int cnt = 1;
  int ssum = sum;
  while (ssum - p[cnt].y > 0) ssum -= p[cnt++].y;
  memset(d, -1, sizeof d);
  for (int i = 1; i <= n; i++) {
    for (int j = i; j >= 2; j--)
      for (int k = aans; k >= 1; k--) {
        if (d[j - 1][k] != -1) d[j][k] = d[j - 1][k];
        if (k >= p[i].y && d[j - 1][k - p[i].y] != -1)
          d[j][k] = max(d[j - 1][k - p[i].y] + p[i].x, d[j][k]);
      }
    d[1][p[i].y] = max(d[1][p[i].y], p[i].x);
  }
  int ans = 0;
  for (int i = aans; i >= sum; i--) {
    if (d[cnt][i] > ans) ans = d[cnt][i];
  }
  cout << cnt << ' ' << sum - ans << endl;
}
