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 long long mod = 998244353;
const int maxn = 105;
struct node {
  int a, b;
} NODE[maxn];
int B[maxn], dp[maxn][maxn * maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int sum = 0;
  for (int i = 1; i <= n; i++) cin >> NODE[i].a, sum += NODE[i].a;
  for (int i = 1; i <= n; i++) cin >> NODE[i].b, B[i] = NODE[i].b;
  sort(B + 1, B + n + 1);
  int Max = 0, idx = n, Msum = 0;
  int sum1 = sum;
  while (sum1 > 0) Max += 1, sum1 -= B[idx], Msum += B[idx], idx--;
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = min(i, Max); j >= 1; j--) {
      for (int k = Msum; k >= NODE[i].b; k--) {
        if (dp[j - 1][k - NODE[i].b] == -1) continue;
        dp[j][k] = max(dp[j - 1][k - NODE[i].b] + NODE[i].a, dp[j][k]);
      }
    }
  }
  int res = 0;
  for (int i = sum; i <= Msum; i++) res = max(res, dp[Max][i]);
  cout << Max << " " << sum - res << endl;
  return 0;
}
