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, K;
struct Bott {
  int a, b;
} p[105];
bool cmp(Bott x, Bott y) {
  if (x.b == y.b) return x.a < y.a;
  return x.b < y.b;
}
int s[105];
int f[2][105][20000 + 10];
int Ans = 0x3f3f3f3f;
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &p[i].a);
  for (int i = 1; i <= N; i++) scanf("%d", &p[i].b);
  sort(p + 1, p + 1 + N, cmp);
  for (int i = 1; i <= N; i++) {
    s[i] = s[i - 1] + p[i].a;
    p[i].b -= p[i].a;
  }
  for (int i = N, sum = 0; i >= 1; i--) {
    sum += p[i].b;
    if (sum >= s[i - 1]) {
      K = N - i + 1;
      break;
    }
  }
  memset(f, 0x3f, sizeof(f));
  f[0][0][10000] = 0;
  int S1 = 10000, S2 = 10000;
  for (int i = 1; i <= N; i++) {
    for (int j = 0; j <= i; j++)
      for (int k = S1; k <= S2; k++) {
        f[i & 1][j + 1][k + p[i].b] =
            min(f[i & 1][j + 1][k + p[i].b], f[(i - 1) & 1][j][k]);
        f[i & 1][j][k - p[i].a] =
            min(f[i & 1][j][k - p[i].a], f[(i - 1) & 1][j][k] + p[i].a);
        f[(i - 1) & 1][j][k] = 0x3f3f3f3f;
      }
    S1 -= p[i].a;
    S2 += p[i].b;
  }
  for (int i = 10000; i <= S2; i++) Ans = min(Ans, f[N & 1][K][i]);
  printf("%d %d", K, Ans);
  return 0;
}
