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 int MAXN = 105;
int n, sum = 0;
int rem[MAXN], cap[MAXN];
map<int, int> dp[MAXN][MAXN];
int getnbot() {
  int tmp[MAXN], tot = sum;
  for (int i = 1; i <= n; i++) tmp[i] = cap[i];
  sort(tmp + 1, tmp + n + 1);
  for (int i = n; i >= 1; i--) {
    if (tot <= tmp[i]) return n - i + 1;
    tot -= tmp[i];
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> rem[i];
    sum += rem[i];
  }
  for (int i = 1; i <= n; i++) cin >> cap[i];
  int nbot = getnbot();
  map<int, int>::iterator it;
  dp[0][0].insert(make_pair(0, 0));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= nbot; j++) {
      for (it = dp[i][j].begin(); it != dp[i][j].end(); it++) {
        if (j < nbot) {
          int nk = it->first + cap[i + 1] - rem[i + 1];
          if (dp[i + 1][j + 1].find(nk) == dp[i + 1][j + 1].end())
            dp[i + 1][j + 1].insert(make_pair(nk, it->second));
          else
            dp[i + 1][j + 1][nk] = min(dp[i + 1][j + 1][nk], it->second);
        }
        int nk = it->first - rem[i + 1];
        int nw = it->second + rem[i + 1];
        if (dp[i + 1][j].find(nk) == dp[i + 1][j].end())
          dp[i + 1][j].insert(make_pair(nk, nw));
        else
          dp[i + 1][j][nk] = min(dp[i + 1][j][nk], nw);
      }
    }
  }
  cout << nbot << ' ';
  int ans = 0x3f3f3f3f;
  for (it = dp[n][nbot].begin(); it != dp[n][nbot].end(); it++) {
    if (it->first >= 0) ans = min(ans, it->second);
  }
  cout << ans << endl;
  return 0;
}
