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 = 1e2 + 20;
int dp[maxn][maxn * maxn];
int a[maxn], b[maxn];
void work() {
  int n;
  scanf("%d", &n);
  int all_A = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    all_A += a[i];
  }
  int all_B = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &b[i]);
    all_B += b[i];
  }
  memset(dp, -0x3f, sizeof dp);
  dp[0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = n; j >= 1; --j) {
      for (int k = all_B; k >= b[i]; --k) {
        dp[j][k] = max(dp[j][k], dp[j - 1][k - b[i]] + a[i]);
      }
    }
  }
  int ansNeed = (0x3f3f3f3f);
  bool flag = true;
  for (int i = 1; i <= n && flag; ++i) {
    for (int j = 1; j <= all_B; ++j) {
      if (dp[i][j] < 0) continue;
      int lef = j - dp[i][j];
      int need = all_A - dp[i][j];
      if (lef >= need) {
        flag = false;
        ansNeed = min(ansNeed, need);
        flag = false;
      }
    }
    if (!flag) {
      cout << i << " " << ansNeed << endl;
      return;
    }
  }
}
int main() {
  work();
  return 0;
}
