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 a[110];
int b[110];
int mem[2][110][10010];
int main() {
  int n, i, j, k;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (i = 1; i <= n; i++) scanf("%d", &b[i]);
  vector<int> v;
  for (i = 1; i <= n; i++) v.push_back(b[i]);
  sort(v.rbegin(), v.rend());
  int sum = 0;
  for (i = 1; i <= n; i++) sum += a[i];
  int cur = 0;
  int cnt = -1;
  for (i = 0; i < n; i++) {
    cur += v[i];
    if (cur >= sum) {
      cnt = i + 1;
      break;
    }
  }
  assert(cnt != -1);
  memset(mem, -1, sizeof mem);
  mem[0][0][0] = 0;
  int c = 1;
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= cnt; j++) {
      for (k = 0; k <= 10000; k++) {
        mem[c][j][k] = mem[1 - c][j][k];
        if (j - 1 >= 0 && k - b[i] >= 0 && mem[1 - c][j - 1][k - b[i]] != -1)
          mem[c][j][k] = max(mem[c][j][k], mem[1 - c][j - 1][k - b[i]] + a[i]);
      }
    }
    c = 1 - c;
  }
  int res = 0;
  for (i = sum; i <= 10000; i++) res = max(res, mem[1 - c][cnt][i]);
  printf("%d %d\n", cnt, sum - res);
  return 0;
}
