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;
static const int INF = 0x3f3f3f3f;
static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
template <typename T, typename U>
static void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static void amax(T &x, U y) {
  if (x < y) x = y;
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    vector<int> a(n);
    for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
    vector<int> b(n);
    for (int i = 0; i < n; ++i) scanf("%d", &b[i]);
    int S = accumulate(a.begin(), a.end(), 0);
    vector<vector<int> > dp(n + 1, vector<int>(S + 1, INF));
    const int Base = (int)1e5;
    dp[0][0] = 0;
    for (int(i) = 0; (i) < (int)(n); ++(i)) {
      for (int(j) = 0; (j) < (int)(S + 1); ++(j)) {
        int x = dp[i][j];
        if (x == INF) continue;
        amin(dp[i + 1][j], x + a[i]);
        amin(dp[i + 1][min(S, j + b[i])], x + Base);
      }
    }
    int ans = dp[n][S];
    printf("%d %d\n", ans / Base, ans % Base);
  }
  return 0;
}
