#include <bits/stdc++.h>
using namespace std;
typedef struct bottle {
  int left, vol;
} bottle;
bottle b[105];
bool cmp(bottle a, bottle b) {
  if (a.vol != b.vol) return a.vol > b.vol;
  return a.left > b.left;
}
int main() {
  int n;
  cin >> n;
  int dp[105][10005];
  int sum = 0;
  memset(dp, -1, sizeof(dp));
  for (int i = 1; i <= n; ++i) {
    cin >> b[i].left;
    sum += b[i].left;
  }
  for (int i = 1; i <= n; ++i) {
    cin >> b[i].vol;
  }
  int cnt = 0;
  int total = 0;
  sort(b + 1, b + n + 1, cmp);
  for (int i = 1; i <= n; ++i) {
    total += b[i].vol;
    if (total >= sum) {
      cnt = i;
      break;
    }
  }
  dp[0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = sum; (j - b[i].left) >= 0; --j) {
      for (int k = i; k >= 1; --k) {
        if (dp[k - 1][j - b[i].left] != -1)
          dp[k][j] = max(dp[k][j], dp[k - 1][j - b[i].left] + b[i].vol);
      }
    }
  }
  int ans = 0;
  for (int i = sum; i >= 1; --i) {
    if (dp[cnt][i] >= sum) {
      ans = sum - i;
      break;
    }
  }
  cout << cnt << " " << ans;
  return 0;
}

### Prompt

Generate 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;
typedef struct bottle {
  int left, vol;
} bottle;
bottle b[105];
bool cmp(bottle a, bottle b) {
  if (a.vol != b.vol) return a.vol > b.vol;
  return a.left > b.left;
}
int main() {
  int n;
  cin >> n;
  int dp[105][10005];
  int sum = 0;
  memset(dp, -1, sizeof(dp));
  for (int i = 1; i <= n; ++i) {
    cin >> b[i].left;
    sum += b[i].left;
  }
  for (int i = 1; i <= n; ++i) {
    cin >> b[i].vol;
  }
  int cnt = 0;
  int total = 0;
  sort(b + 1, b + n + 1, cmp);
  for (int i = 1; i <= n; ++i) {
    total += b[i].vol;
    if (total >= sum) {
      cnt = i;
      break;
    }
  }
  dp[0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = sum; (j - b[i].left) >= 0; --j) {
      for (int k = i; k >= 1; --k) {
        if (dp[k - 1][j - b[i].left] != -1)
          dp[k][j] = max(dp[k][j], dp[k - 1][j - b[i].left] + b[i].vol);
      }
    }
  }
  int ans = 0;
  for (int i = sum; i >= 1; --i) {
    if (dp[cnt][i] >= sum) {
      ans = sum - i;
      break;
    }
  }
  cout << cnt << " " << ans;
  return 0;
}

```