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;
int n, sum;
int dp[105][10005];
struct node {
  int a, b;
} m[105];
bool cmp(node a, node b) {
  if (a.b == b.b) return a.a < b.a;
  return a.b < b.b;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &m[i].a);
    sum += m[i].a;
  }
  for (int i = 1; i <= n; i++) scanf("%d", &m[i].b);
  sort(m + 1, m + n + 1, cmp);
  int ans = 0;
  int ii;
  for (int i = n; i >= 1; i--) {
    ans += m[i].b;
    if (ans >= sum) {
      ii = i;
      printf("%d ", n - i + 1);
      break;
    }
  }
  int nn = n - ii + 1;
  for (int k = 1; k <= n; k++)
    for (int i = min(k, nn); i >= 1; i--) {
      if (i == 1)
        dp[1][m[k].b] = max(dp[1][m[k].b], m[k].a);
      else
        for (int j = 10000; j >= 0; j--) {
          if (dp[i - 1][j] == 0) continue;
          dp[i][j + m[k].b] = max(dp[i - 1][j] + m[k].a, dp[i][j + m[k].b]);
        }
    }
  int anss = 0;
  for (int i = sum; i <= 10000; i++) {
    if (dp[nn][i] > anss) anss = dp[nn][i];
  }
  printf("%d\n", sum - anss);
  return 0;
}
