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;
template <typename T>
inline T qmin(const T &a, const T &b) {
  return a < b ? a : b;
}
template <typename T>
inline T qmax(const T &a, const T &b) {
  return a > b ? a : b;
}
template <typename T>
inline void getmin(T &a, const T &b) {
  if (a > b) a = b;
}
template <typename T>
inline void getmax(T &a, const T &b) {
  if (a < b) a = b;
}
inline void fileio(string s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
const int inf = (int)1e9 + 7;
const long long linf = (long long)1e17 + 7;
const int N = 101;
int f[N][N * N];
int a[N], b[N], id[N], n, suma, sumb, tot;
inline bool cmp(const int &x, const int &y) { return b[x] > b[y]; }
int main() {
  scanf("%d", &n);
  for (register int i = (1); i <= (n); ++i) scanf("%d", a + i), suma += a[i];
  for (register int i = (1); i <= (n); ++i)
    scanf("%d", b + i), sumb += b[i], id[i] = i;
  sort(id + 1, id + n + 1, cmp);
  for (int rest = suma; rest > 0; rest -= b[id[++tot]])
    ;
  for (register int i = (0); i <= (tot); ++i)
    for (register int j = (0); j <= (sumb); ++j) f[i][j] = -inf;
  f[0][0] = 0;
  for (register int i = (1); i <= (n); ++i) {
    for (register int j = (tot); j >= (1); --j) {
      for (register int k = (b[i]); k <= (sumb); ++k) {
        getmax(f[j][k], f[j - 1][k - b[i]] + a[i]);
      }
    }
  }
  int ans = 0;
  for (register int k = (suma); k <= (sumb); ++k) getmax(ans, f[tot][k]);
  printf("%d %d\n", tot, suma - ans);
  return 0;
}
