#include <bits/stdc++.h>
using namespace std;
int a[105], b[105];
pair<int, int> dp[105][10000 * 2 + 5];
int n;
pair<int, int> rek(int p, int left) {
  if (p == n) {
    if (left >= 10000) return pair<int, int>(0, 0);
    return pair<int, int>(1000, 1000000);
  }
  pair<int, int> &res = dp[p][left];
  if (res.first != -1) return res;
  res = pair<int, int>(1000, 100000);
  pair<int, int> temp;
  temp = rek(p + 1, left - a[p]);
  temp.second += a[p];
  if (temp.first < res.first ||
      (temp.first == res.first && temp.second < res.second))
    res = temp;
  temp = rek(p + 1, left + (b[p] - a[p]));
  temp.first += 1;
  if (temp.first < res.first ||
      (temp.first == res.first && temp.second < res.second))
    res = temp;
  return res;
}
int main() {
  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++) {
    for (int j = 0; j <= 20000; j++) {
      dp[i][j] = pair<int, int>(-1, -1);
    }
  }
  pair<int, int> res = rek(0, 10000);
  cout << res.first << " " << res.second << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
int a[105], b[105];
pair<int, int> dp[105][10000 * 2 + 5];
int n;
pair<int, int> rek(int p, int left) {
  if (p == n) {
    if (left >= 10000) return pair<int, int>(0, 0);
    return pair<int, int>(1000, 1000000);
  }
  pair<int, int> &res = dp[p][left];
  if (res.first != -1) return res;
  res = pair<int, int>(1000, 100000);
  pair<int, int> temp;
  temp = rek(p + 1, left - a[p]);
  temp.second += a[p];
  if (temp.first < res.first ||
      (temp.first == res.first && temp.second < res.second))
    res = temp;
  temp = rek(p + 1, left + (b[p] - a[p]));
  temp.first += 1;
  if (temp.first < res.first ||
      (temp.first == res.first && temp.second < res.second))
    res = temp;
  return res;
}
int main() {
  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++) {
    for (int j = 0; j <= 20000; j++) {
      dp[i][j] = pair<int, int>(-1, -1);
    }
  }
  pair<int, int> res = rek(0, 10000);
  cout << res.first << " " << res.second << endl;
  return 0;
}

```