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;
int n, k, dp[105][10005];
struct node {
  int now, lim;
};
node nod[105];
bool nodecmp(node x, node y) { return x.lim > y.lim; }
using namespace std;
int main() {
  cin >> n;
  int water = 0, maxx = 0;
  for (int i = 1; i <= n; i++) {
    cin >> nod[i].now;
    water += nod[i].now;
  }
  for (int i = 1; i <= n; i++) {
    cin >> nod[i].lim;
    maxx += nod[i].lim;
  }
  sort(nod + 1, nod + n + 1, nodecmp);
  int k = 0;
  int temp = water;
  while (temp > 0) {
    k++;
    temp -= nod[k].lim;
  }
  cout << k << " ";
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int x = maxx; x >= nod[i].lim; x--) {
      for (int j = 1; j <= k; j++) {
        if (dp[j - 1][x - nod[i].lim] != -1) {
          dp[j][x] = max(dp[j][x], dp[j - 1][x - nod[i].lim] + nod[i].now);
        }
      }
    }
  }
  int ans = 0;
  for (int i = water; i <= maxx; i++) {
    ans = max(ans, dp[k][i]);
  }
  cout << water - ans << endl;
}
