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;
const int N = 101, INF = 0x3f3f3f3f;
int a[N], b[N], dp[2][N][2 * N * N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  memset(dp, 0x3f, sizeof dp);
  for (int i = N * N; i < 2 * N * N; i++) dp[0][0][i] = 0;
  int q = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      for (int k = 0; k < 2 * N * N; k++) {
        dp[q][j][k] = min(
            k >= a[i] ? a[i] + dp[q ^ 1][j][k - a[i]] : INF,
            j && k + b[i] >= a[i] ? dp[q ^ 1][j - 1][k + b[i] - a[i]] : INF);
      }
    }
    q ^= 1;
  }
  for (int i = 1; i <= n; i++) {
    if (dp[q ^ 1][i][N * N] < INF) {
      printf("%d %d\n", i, dp[q ^ 1][i][N * N]);
      return 0;
    }
  }
}
