#include <bits/stdc++.h>
using namespace std;
const int N = 110;
int botlsiz[N], arr[N];
map<int, pair<int, int>> dp[N];
int n;
pair<int, int> solve(int i, int c) {
  if (i == n)
    if (c >= 0)
      return {0, 0};
    else
      return {1e6, 1e6};
  if (dp[i].count(c)) return dp[i][c];
  auto ans = solve(i + 1, c + arr[i]);
  ans.first += 1;
  auto tmp = solve(i + 1, c - botlsiz[i]);
  tmp.second += botlsiz[i];
  ans = min(ans, tmp);
  return dp[i][c] = ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", botlsiz + i);
  for (int i = 0; i < n; i++) {
    scanf("%d", arr + i);
    arr[i] -= botlsiz[i];
  }
  auto ans = solve(0, 0);
  printf("%d %d\n", ans.first, ans.second);
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 110;
int botlsiz[N], arr[N];
map<int, pair<int, int>> dp[N];
int n;
pair<int, int> solve(int i, int c) {
  if (i == n)
    if (c >= 0)
      return {0, 0};
    else
      return {1e6, 1e6};
  if (dp[i].count(c)) return dp[i][c];
  auto ans = solve(i + 1, c + arr[i]);
  ans.first += 1;
  auto tmp = solve(i + 1, c - botlsiz[i]);
  tmp.second += botlsiz[i];
  ans = min(ans, tmp);
  return dp[i][c] = ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", botlsiz + i);
  for (int i = 0; i < n; i++) {
    scanf("%d", arr + i);
    arr[i] -= botlsiz[i];
  }
  auto ans = solve(0, 0);
  printf("%d %d\n", ans.first, ans.second);
  return 0;
}

```