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 dp[2][110][110 * 110];
int a[110], b[110], c[110];
template <typename T>
void smax(T& x, T y) {
  if (x < y) x = y;
}
int main() {
  int N, S, K, SS, tp, ans;
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", a + i);
  for (int i = 0; i < N; i++) scanf("%d", b + i);
  for (int i = 0; i < N; i++) c[i] = b[i];
  sort(c, c + N);
  reverse(c, c + N);
  S = 0;
  for (int i = 0; i < N; i++) S += a[i];
  for (int i = 0; i < N; i++) {
    S -= c[i];
    if (S <= 0) {
      K = i + 1;
      break;
    }
  }
  S = 0;
  for (int i = 0; i < N; i++) S += a[i];
  SS = 0;
  for (int i = 0; i < N; i++) SS += b[i];
  tp = 0;
  memset(dp[tp], 0xff, sizeof(dp[tp]));
  dp[tp][0][0] = 0;
  dp[tp][1][b[0]] = a[0];
  for (int i = 1; i < N; i++) {
    for (int k = 0; k <= K; k++)
      for (int j = 0; j <= SS; j++) {
        dp[tp ^ 1][k][j] = -1;
        if (k == 0 && j == 0) {
          dp[tp ^ 1][k][j] = 0;
          continue;
        }
        dp[tp ^ 1][k][j] = dp[tp][k][j];
        if (k && j >= b[i] && dp[tp][k - 1][j - b[i]] != -1) {
          smax(dp[tp ^ 1][k][j], dp[tp][k - 1][j - b[i]] + a[i]);
        }
      }
    tp = tp ^ 1;
  }
  ans = 0;
  for (int i = S; i <= SS; i++) smax(ans, dp[tp][K][i]);
  printf("%d %d\n", K, S - ans);
  return 0;
}
