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>
#pragma gcc optimize("O3")
#pragma gcc optimize("unroll-loops")
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const int N = 110;
const int INF = 1e9;
int n, tot;
pii ar[N];
int dp[N][N * N];
int ndp[N][N * N];
int findk() {
  int ret = 0;
  int tot = 0;
  for (int i = 1; i <= n; i++) tot += ar[i].second;
  for (int i = 1; i <= n; i++) {
    tot -= ar[i].first;
    if (tot <= 0) return i;
  }
  return n;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ar[i].second;
    tot += ar[i].second;
  }
  for (int i = 1; i <= n; i++) {
    cin >> ar[i].first;
  }
  sort(ar + 1, ar + n + 1, greater<pii>());
  int mk = findk();
  for (int i = 0; i <= mk; i++) {
    for (int j = 0; j <= tot; j++) {
      dp[i][j] = INF;
    }
  }
  dp[0][0] = tot;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= mk; j++) {
      for (int k = 0; k <= tot; k++) {
        ndp[j][k] = dp[j][k];
      }
    }
    for (int j = 0; j < mk; j++) {
      for (int k = 0; k <= tot; k++) {
        int nj = j + 1;
        int nk = min(tot, k + ar[i].first);
        dp[nj][nk] = min(dp[nj][nk], ndp[j][k] - ar[i].second);
      }
    }
  }
  cout << mk << " " << dp[mk][tot] << '\n';
  return 0;
}
