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;
  bool operator<(const node& r) const { return r.b < b; }
} e[105];
int dp[10005][105];
int main() {
  int n, sum = 0, num = 0, sum2 = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &e[i].a);
    sum += e[i].a;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &e[i].b);
    sum2 += e[i].b;
  }
  sort(e, e + n);
  for (int i = 0; i < n; i++) {
    num += e[i].b;
    if (num >= sum) {
      num = i + 1;
      break;
    }
  }
  for (int i = sum2; i >= 0; i--) {
    for (int k = num; k >= 0; k--) {
      dp[i][k] = -234723846;
    }
  }
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int k = sum2; k >= e[i].b; k--) {
      for (int j = num; j >= 1; j--) {
        dp[k][j] = max(dp[k][j], dp[k - e[i].b][j - 1] + e[i].a);
      }
    }
  }
  int ans = 0;
  for (int i = sum; i <= sum2; i++) {
    ans = max(ans, dp[i][num]);
  }
  printf("%d %d\n", num, sum - ans);
}
