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 dp[105][10005];
struct node {
  int a, b;
} bottle[105];
bool cmp(node t1, node t2) {
  if (t1.b != t2.b)
    return t1.b > t2.b;
  else
    return t1.a > t2.a;
}
int main() {
  int n, sum = 0;
  int cnt;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &bottle[i].a);
    sum += bottle[i].a;
  }
  for (int i = 1; i <= n; i++) scanf("%d", &bottle[i].b);
  sort(bottle + 1, bottle + n + 1, cmp);
  for (int i = 1, tmp = 0; i <= n; i++) {
    tmp += bottle[i].b;
    if (sum <= tmp) {
      cnt = i;
      break;
    }
  }
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = sum - bottle[i].a; j >= 0; j--) {
      for (int pre = i - 1; pre >= 0; pre--) {
        if (dp[pre][j] != -1) {
          dp[pre + 1][j + bottle[i].a] =
              max(dp[pre + 1][j + bottle[i].a], dp[pre][j] + bottle[i].b);
        }
      }
    }
  }
  int t;
  for (int i = sum; i >= 0; i--) {
    if (dp[cnt][i] >= sum) {
      t = i;
      break;
    }
  }
  printf("%d %d\n", cnt, sum - t);
  return 0;
}
