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 x, y;
} s[105];
int cmp(node a, node b) { return a.y > b.y; }
int i, n, k, t, j, m, sum = 0, ans = 0;
int dp[105][100005];
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &s[i].x);
    sum += s[i].x;
  }
  for (i = 1; i <= n; i++) scanf("%d", &s[i].y);
  sort(s + 1, s + n + 1, cmp);
  for (i = 1; i <= n; i++) {
    ans += s[i].y;
    if (ans >= sum) {
      k = i;
      break;
    }
  }
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (i = 1; i <= n; i++) {
    for (j = sum - s[i].x; j >= 0; j--) {
      for (m = i - 1; m >= 0; m--) {
        if (dp[m][j] != -1)
          dp[m + 1][j + s[i].x] = max(dp[m + 1][j + s[i].x], dp[m][j] + s[i].y);
      }
    }
  }
  for (i = sum; i >= 0; i--) {
    if (dp[k][i] >= sum) {
      t = i;
      break;
    }
  }
  printf("%d %d\n", k, sum - t);
}
