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;
vector<vector<vector<int>>> dp;
vector<int> p;
int check(vector<pair<int, int>>& a, int curr, int capa, int count, int mini,
          int sum) {
  if (dp[curr][count][capa] != 1000000000) return dp[curr][count][capa];
  if (curr == a.size() && count == mini && capa >= sum)
    return dp[curr][count][capa] = 0;
  else if (curr == a.size())
    return dp[curr][count][capa] = 1000000000;
  if (count <= mini && curr + mini - count <= a.size() &&
      (capa + p[curr] - p[curr + (mini - count)]) < sum)
    return dp[curr][count][capa] = 1000000000;
  int res = 1000000000;
  res = min(res, check(a, curr + 1, capa, count, mini, sum) + a[curr].second);
  if (count < mini)
    res = min(res,
              check(a, curr + 1, capa + a[curr].first, count + 1, mini, sum));
  return dp[curr][count][capa] = res;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; i++) cin >> a[i].second;
  for (int i = 0; i < n; i++) cin >> a[i].first;
  sort(a.begin(), a.end(), greater<pair<int, int>>());
  int sum = 0;
  for (int i = 0; i < n; i++) {
    sum += a[i].second;
  }
  int orig = sum;
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (sum - a[i].first <= 0) {
      cnt = i + 1;
      break;
    }
    sum -= a[i].first;
  }
  dp.assign(n + 1, vector<vector<int>>(n + 1, vector<int>(10500, 1000000000)));
  int ans = 1000000000;
  dp[0][0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= cnt; j++) {
      for (int k = 0; k <= cnt * 101; k++) {
        if (dp[i][j][k] == 1000000000) continue;
        dp[i + 1][j][k] = min(dp[i + 1][j][k], dp[i][j][k] + a[i].second);
        dp[i + 1][j + 1][k + a[i].first] =
            min(dp[i + 1][j + 1][k + a[i].first], dp[i][j][k]);
      }
    }
  }
  for (int i = orig; i <= cnt * 101; i++) {
    ans = min(ans, dp[n][cnt][i]);
  }
  cout << cnt << " " << ans << endl;
  return 0;
}
