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;
long long mem[101][10010], t[101][10010];
long long n, a[101], b[101], total;
long long dp(long long cur, long long ob) {
  if (ob == 0) {
    t[cur][ob] = 0;
    return 0;
  }
  if (n == cur) {
    t[cur][ob] = 999999;
    return 200;
  }
  long long &ret = mem[cur][ob];
  if (ret != -1) return ret;
  long long cob, x;
  if (ob < b[cur]) {
    x = ob - a[cur];
    cob = 0;
  } else {
    cob = ob - b[cur];
    x = b[cur] - a[cur];
  }
  ret = 1 + dp(cur + 1, cob);
  t[cur][ob] = x + t[cur + 1][cob];
  if (ret == dp(cur + 1, ob)) {
    t[cur][ob] = min(t[cur][ob], t[cur + 1][ob]);
  } else if (ret > dp(cur + 1, ob)) {
    ret = dp(cur + 1, ob);
    t[cur][ob] = t[cur + 1][ob];
  }
  return ret;
}
int main() {
  long long total = 0;
  memset(mem, -1, sizeof(mem));
  scanf("%lld", &n);
  for (long long i = 0; i < n; i++) {
    scanf("%lld", a + i);
    total += a[i];
  }
  for (long long i = 0; i < n; i++) scanf("%lld", b + i);
  cout << dp(0, total);
  printf(" ");
  cout << t[0][total] << endl;
  return 0;
}
