#include <bits/stdc++.h>
using namespace std;
const int N = 110, M = 10010;
int n;
int a[N], b[N], c[N];
int f[M][N];
int main() {
  cin >> n;
  int sa = 0, sb = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sa += a[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    c[i] = b[i];
    sb += b[i];
  }
  sort(c + 1, c + n + 1, greater<int>());
  int s = 0, cnt = 0;
  for (int i = 1; i <= n; i++) {
    s += c[i];
    cnt++;
    if (s >= sa) break;
  }
  memset(f, -0x3f, sizeof f);
  f[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = sb; j >= b[i]; j--) {
      for (int k = cnt; k >= 1; k--)
        f[j][k] = max(f[j][k], f[j - b[i]][k - 1] + a[i]);
    }
  }
  int res = 0;
  for (int i = sa; i <= sb; i++) res = max(res, f[i][cnt]);
  res = sa - res;
  cout << cnt << " " << res << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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, M = 10010;
int n;
int a[N], b[N], c[N];
int f[M][N];
int main() {
  cin >> n;
  int sa = 0, sb = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sa += a[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    c[i] = b[i];
    sb += b[i];
  }
  sort(c + 1, c + n + 1, greater<int>());
  int s = 0, cnt = 0;
  for (int i = 1; i <= n; i++) {
    s += c[i];
    cnt++;
    if (s >= sa) break;
  }
  memset(f, -0x3f, sizeof f);
  f[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = sb; j >= b[i]; j--) {
      for (int k = cnt; k >= 1; k--)
        f[j][k] = max(f[j][k], f[j - b[i]][k - 1] + a[i]);
    }
  }
  int res = 0;
  for (int i = sa; i <= sb; i++) res = max(res, f[i][cnt]);
  res = sa - res;
  cout << cnt << " " << res << endl;
  return 0;
}

```