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;
bool cmp(pair<int, int> a, pair<int, int> b) { return a.second > b.second; }
pair<int, int> a[105];
int dp2[105][10005];
int dp[105][10005], tot, n;
int solve(int i, int j) {
  if (j < 0) return 100000000;
  if (i == 0) {
    if (j > a[i].second) return dp[i][j] = 100000000;
    dp2[i][j] = (j && 1);
    return dp[i][j] = abs(j - a[i].first);
  }
  if (dp[i][j] != -1) return dp[i][j];
  dp[i][j] = 100000000;
  dp2[i][j] = 100000000;
  for (int x = 0; x <= a[i].second && x <= j; ++x) {
    int tmp = abs(x - a[i].first) + solve(i - 1, j - x);
    if (dp2[i - 1][j - x] == -1) continue;
    if (dp2[i - 1][j - x] + (x && 1) < dp2[i][j]) {
      dp[i][j] = tmp;
      dp2[i][j] = dp2[i - 1][j - x] + (x && 1);
      continue;
    }
    if (dp2[i - 1][j - x] + (x && 1) == dp2[i][j]) {
      dp[i][j] = min(dp[i][j], tmp);
      continue;
    }
  }
  return dp[i][j];
}
int main() {
  for (int i = 0; i < 105; ++i) {
    for (int j = 0; j < 10005; ++j) {
      dp[i][j] = -1;
      dp2[i][j] = -1;
    }
  }
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i].first);
    tot += a[i].first;
  }
  for (int i = 0; i < n; ++i) scanf("%d", &a[i].second);
  sort(a, a + n, cmp);
  solve(n - 1, tot);
  printf("%d %d\n", dp2[n - 1][tot], dp[n - 1][tot] / 2);
  return 0;
}
