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 = (1 << 17) + 10;
const int inf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
int n;
struct NODE {
  int a, b;
} node[110];
bool cmp(NODE x, NODE y) { return x.b > y.b; }
int dp[2][110][10010], sum[110];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  int tot = 0, cnt = 0, tmp = 0;
  for (int i = 1; i <= n; i++) {
    cin >> node[i].a;
    tot += node[i].a;
  }
  for (int i = 1; i <= n; i++) {
    cin >> node[i].b;
  }
  if (n == 1) {
    cout << 1 << " " << 0 << endl;
    return 0;
  }
  sort(node + 1, node + 1 + n, cmp);
  for (int i = 1; i <= n; i++) {
    tmp += node[i].b;
    cnt++;
    if (tmp >= tot) break;
  }
  sum[0] = 0;
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + node[i].a;
  }
  memset(dp, -inf, sizeof(dp));
  dp[0][0][0] = dp[1][0][0] = 0;
  int t = 0;
  for (int i = 1; i <= n; i++, t = t ^ 1) {
    for (int j = 1; j <= i && j <= cnt; j++) {
      for (int k = 0; k <= sum[i]; k++) {
        dp[t][j][k] = dp[t ^ 1][j][k];
        if (k >= node[i].a) {
          dp[t][j][k] =
              max(dp[t][j][k], dp[t ^ 1][j - 1][k - node[i].a] + node[i].b);
        }
      }
    }
  }
  for (int i = tot; i >= 0; i--) {
    if (dp[t ^ 1][cnt][i] >= tot) {
      cout << cnt << " " << tot - i << endl;
      break;
    }
  }
  return 0;
}
