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("Ofast,unroll-loops")
using namespace std;
const int N = 102;
const int oo = 1e9;
const int Mod = 1e9 + 7;
int dp[2][N][N * N];
int a[N], b[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (int i = (int)(0); i < (int)(2); ++i)
    for (int j = (int)(0); j < (int)(102); ++j)
      for (int k = (int)(0); k < (int)(10002); ++k) dp[i][j][k] = -oo;
  dp[0][0][0] = 0;
  int n;
  cin >> n;
  vector<int> caps;
  int vtot = 0, ctot = 0;
  for (int i = (int)(1); i < (int)(n + 1); ++i) cin >> a[i], vtot += a[i];
  for (int i = (int)(1); i < (int)(n + 1); ++i)
    cin >> b[i], caps.push_back(b[i]), ctot += b[i];
  sort(caps.begin(), caps.end(), greater<int>());
  int k = 0, acc = 0;
  while (acc < vtot) acc += caps[k++];
  for (int i = 1; i <= n; ++i) {
    int idx = i % 2;
    for (int j = 0; j <= k; ++j) {
      for (int c = 0; c <= ctot; ++c) {
        if (j >= 1 and c - b[i] >= 0)
          dp[idx][j][c] =
              max(dp[idx][j][c], dp[idx ^ 1][j - 1][c - b[i]] + a[i]);
        dp[idx][j][c] = max(dp[idx][j][c], dp[idx ^ 1][j][c]);
      }
    }
  }
  int idx = n % 2;
  int high = 0;
  for (int c = vtot; c <= ctot; ++c) {
    int soda = dp[idx][k][c];
    high = max(soda, high);
  }
  cout << k << " " << vtot - high << '\n';
  return 0;
}
