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;
struct Bottle {
  int water, all;
  inline bool operator<(const Bottle &x) const { return all > x.all; }
};
int n, ans, sum, now, dp[10005][105], ans_time = -2147483647;
Bottle bottle[105];
inline int read() {
  int ans = 0, f = 1;
  char in = getchar();
  while (!(in >= '0' && in <= '9')) {
    if (in == '-') {
      f = !f;
    }
    in = getchar();
  }
  while (in >= '0' && in <= '9') {
    ans = (ans << 1) + (ans << 3) + (in ^ '0');
    in = getchar();
  }
  return ans * f;
}
int main() {
  n = read();
  for (register int i = 1; i <= n; i++) {
    bottle[i].water = read();
    sum += bottle[i].water;
  }
  for (register int i = 1; i <= n; i++) {
    bottle[i].all = read();
  }
  sort(bottle + 1, bottle + n + 1);
  while (now < sum) {
    ans++;
    now += bottle[ans].all;
  }
  printf("%d ", ans);
  memset(dp, 128, sizeof(dp));
  dp[0][0] = 0;
  for (register int i = 1; i <= n; i++) {
    for (register int j = now; j >= bottle[i].all; j--) {
      for (register int k = 1; k <= ans; k++) {
        dp[j][k] = (dp[j][k] > dp[j - bottle[i].all][k - 1] + bottle[i].water
                        ? dp[j][k]
                        : dp[j - bottle[i].all][k - 1] + bottle[i].water);
      }
    }
  }
  for (register int i = sum; i <= now; i++) {
    ans_time = (ans_time > dp[i][ans] ? ans_time : dp[i][ans]);
  }
  printf("%d\n", sum - ans_time);
  return 0;
}
