#include <bits/stdc++.h>
using namespace std;
const int N = 1e2 + 10, M = 2e4 + 10;
int dp[M][N], a[N], b[N];
int main() {
  int n, sum = 0, mx = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    mx += b[i];
  }
  mx += sum;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= mx + 2; j++) dp[j][i] = -1e9;
  for (int i = 0; i <= n; i++) dp[0][i] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = mx; j >= 0; j--) {
      if (j - b[i] < 0) break;
      for (int k = 1; k <= n; k++) {
        if (dp[j - b[i]][k - 1] >= 0)
          dp[j][k] = max(dp[j][k], dp[j - b[i]][k - 1] + a[i]);
      }
    }
  }
  for (int i = mx; i >= 0; i--)
    for (int j = 0; j <= n; j++) dp[i][j] = max(dp[i][j], dp[i + 1][j]);
  for (int i = 0; i <= n; i++)
    if (dp[sum][i] >= 0) {
      cout << i << " " << sum - dp[sum][i] << "\n";
      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 = 1e2 + 10, M = 2e4 + 10;
int dp[M][N], a[N], b[N];
int main() {
  int n, sum = 0, mx = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    mx += b[i];
  }
  mx += sum;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= mx + 2; j++) dp[j][i] = -1e9;
  for (int i = 0; i <= n; i++) dp[0][i] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = mx; j >= 0; j--) {
      if (j - b[i] < 0) break;
      for (int k = 1; k <= n; k++) {
        if (dp[j - b[i]][k - 1] >= 0)
          dp[j][k] = max(dp[j][k], dp[j - b[i]][k - 1] + a[i]);
      }
    }
  }
  for (int i = mx; i >= 0; i--)
    for (int j = 0; j <= n; j++) dp[i][j] = max(dp[i][j], dp[i + 1][j]);
  for (int i = 0; i <= n; i++)
    if (dp[sum][i] >= 0) {
      cout << i << " " << sum - dp[sum][i] << "\n";
      return 0;
    }
}

```