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;
struct Node {
  int a, b;
} x[110];
bool cmp(Node i, Node j) {
  if (i.a == j.a)
    return i.b > j.b;
  else
    return i.a > j.a;
}
int n, ans, sum, k, dp[110][10010], sumv;
int main() {
  while (~scanf("%d", &n)) {
    sumv = ans = sum = k = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &x[i].b);
      sum += x[i].b;
    }
    for (int i = 0; i < n; i++) {
      scanf("%d", &x[i].a);
      sumv += x[i].a;
    }
    sort(x, x + n, cmp);
    int temp = sum, cnt = 0;
    while (temp > 0) {
      temp -= x[cnt++].a;
      k++;
    }
    memset(dp, -0x3f3f3f3f, sizeof(dp));
    dp[0][0] = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = sumv; j >= x[i - 1].a; j--) {
        for (int u = 1; u <= k; u++) {
          dp[u][j] = max(dp[u][j], dp[u - 1][j - x[i - 1].a] + x[i - 1].b);
        }
      }
    }
    ans = -0x3f3f3f3f;
    for (int i = sum; i <= sumv; i++) ans = max(ans, dp[k][i]);
    printf("%d %d\n", k, sum - ans);
  }
}
