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 double pi = acos(-1.0);
const long double eps = 1e-10;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MX = 100 + 7;
const int mod = 1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long quimod(long long a, long long k) {
  long long res = 1;
  while (k) {
    if (k & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    k >>= 1;
  }
  return res;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
struct node {
  int r, v;
  inline bool operator<(const node& w) const { return v > w.v; }
} a[MX];
int dp[MX][MX * MX];
int main() {
  memset(dp, -1, sizeof(dp));
  ;
  int n;
  scanf("%d", &n);
  int tot_r = 0, tot_v = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i].r);
    tot_r += a[i].r;
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i].v);
    tot_v += a[i].v;
  }
  sort(a + 1, a + 1 + n);
  int now = 0;
  int num = 0;
  for (int i = 1; i <= n; ++i) {
    now += a[i].v;
    if (now >= tot_r) {
      num = i;
      break;
    }
  }
  dp[0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = num; j > 0; --j) {
      for (int k = tot_v; k >= a[i].v; --k) {
        if (dp[j - 1][k - a[i].v] != -1)
          dp[j][k] = max(dp[j][k], dp[j - 1][k - a[i].v] + a[i].r);
      }
    }
  }
  int res = 0;
  for (int i = tot_r; i <= tot_v; ++i) {
    res = max(res, dp[num][i]);
  }
  printf("%d %d\n", num, tot_r - res);
  return 0;
}
