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 long long INF = 0x7f7f7f7f;
long long n, m, a[105], b[105];
long long ans1 = INF, ans2 = INF;
long long f[10005], g[10005];
signed main() {
  memset(f, INF, sizeof(f));
  cin >> n;
  for (long long i = 1; i <= n; cin >> a[i++])
    ;
  for (long long i = 1; i <= n; cin >> b[i++])
    ;
  for (long long i = 1; i <= n; i++) m += b[i];
  f[0] = 0;
  for (long long i = 1; i <= n; i++)
    for (long long j = m; j >= b[i]; j--)
      if (f[j - b[i]] + 1 < f[j])
        f[j] = f[j - b[i]] + 1, g[j] = g[j - b[i]] + a[i];
      else if (f[j - b[i]] + 1 == f[j] && g[j - b[i]] + a[i] > g[j])
        g[j] = g[j - b[i]] + a[i];
  long long sum = 0;
  for (long long i = 1; i <= n; i++) sum += a[i];
  for (long long i = sum; i <= m; i++)
    if (f[i] < ans1)
      ans1 = f[i], ans2 = g[i];
    else if (f[i] == ans1 && g[i] > ans2)
      ans2 = g[i];
  cout << ans1 << ' ' << sum - ans2 << endl;
}
