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;
const int maxn = 101;
int n, a[maxn], b[maxn], f[maxn][10000], water = 0, ans = 0, m, q, w = 0,
                                         g[maxn][10000];
priority_queue<int> que;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    water += a[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    que.push(b[i]);
    w += b[i];
  }
  int tot = 0;
  while (tot < water) {
    ans++;
    int x = que.top();
    que.pop();
    tot += x;
  }
  printf("%d ", ans);
  m = n - ans;
  q = w - water;
  memset(g, 63, sizeof(g));
  for (int i = 0; i <= q; i++) g[0][i] = 0;
  for (int i = b[1]; i <= q; i++) g[1][i] = a[1];
  for (int i = 2; i <= n; i++) {
    memset(f, 63, sizeof(f));
    for (int j = 0; j <= q; j++) {
      f[0][j] = 0;
    }
    for (int j = 1; j <= m; j++) {
      if (j > i) continue;
      for (int k = 0; k <= q; k++) {
        f[j][k] = g[j][k];
        if (k >= b[i]) f[j][k] = min(f[j][k], g[j - 1][k - b[i]] + a[i]);
      }
    }
    for (int j = 0; j <= m; j++)
      for (int k = 0; k <= q; k++) g[j][k] = f[j][k];
  }
  printf("%d\n", g[m][q]);
  return 0;
}
