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 Set(long long n, long long pos) { return (n = n | 1 << pos); }
long long reset(long long N, long long pos) { return N = N & ~(1 << pos); }
bool check(long long N, long long pos) { return (bool)(N & (1 << pos)); }
long long pow(long long a, long long b, long long mod) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > mod) x %= mod;
    }
    y = (y * y);
    if (y > mod) y %= mod;
    b /= 2;
  }
  return x;
}
long long modInverse(long long a, long long m) { return pow(a, m - 2, m); }
const long long LARGE = powl(10, 18);
long long n;
long long arr[201], brr[201];
long long total = 0;
pair<long long, long long> dp[201][20100];
bool vis[201][20100];
pair<long long, long long> fun(long long idx, long long have) {
  if (idx == n) {
    if (have >= total) {
      return make_pair(0, 0);
    } else
      return make_pair(INT_MAX, INT_MAX);
  }
  if (vis[idx][have] != 0) return dp[idx][have];
  vis[idx][have] = 1;
  pair<long long, long long> ans = make_pair(INT_MAX, INT_MAX);
  ans = min(ans, make_pair(1 + fun(idx + 1, have + brr[idx]).first,
                           -arr[idx] + fun(idx + 1, have + brr[idx]).second));
  ans = min(ans, fun(idx + 1, have));
  return dp[idx][have] = ans;
}
int main() {
  while (cin >> n) {
    total = 0;
    memset(vis, 0, sizeof vis);
    for (int i = 0; i < n; i++) cin >> arr[i], total += arr[i];
    for (int i = 0; i < n; i++) cin >> brr[i];
    pair<long long, long long> ans = fun(0, 0);
    cout << ans.first << " " << total + ans.second << endl;
  }
}
