#include <bits/stdc++.h>
using namespace std;
int dp[106][10006];
struct bottle {
  int remain, volume;
} f[105];
bool cmp(bottle a, bottle b) {
  if (a.volume != b.volume)
    return a.volume > b.volume;
  else
    return a.remain > b.remain;
}
int main() {
  int n;
  cin >> n;
  int sum1, sum2;
  sum1 = sum2 = 0;
  for (int i = 1; i <= n; i++) {
    cin >> f[i].remain;
    sum1 += f[i].remain;
  }
  for (int i = 1; i <= n; i++) {
    cin >> f[i].volume;
    sum2 += f[i].volume;
  }
  int x = sum1;
  int num = 0;
  sort(f + 1, f + 1 + n, cmp);
  while (1) {
    x -= f[++num].volume;
    if (x <= 0) break;
  }
  for (int i = 0; i <= 105; i++)
    for (int j = 0; j <= 10005; j++) dp[i][j] = -999999;
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = sum2; j >= f[i].volume; j--)
      for (int k = 1; k <= num; k++)
        dp[k][j] = max(dp[k][j], dp[k - 1][j - f[i].volume] + f[i].remain);
  }
  int ans = 0;
  for (int i = sum1; i <= sum2; i++) {
    ans = max(ans, dp[num][i]);
  }
  printf("%d %d\n", num, sum1 - ans);
}

### Prompt

Please create a solution in Cpp 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;
int dp[106][10006];
struct bottle {
  int remain, volume;
} f[105];
bool cmp(bottle a, bottle b) {
  if (a.volume != b.volume)
    return a.volume > b.volume;
  else
    return a.remain > b.remain;
}
int main() {
  int n;
  cin >> n;
  int sum1, sum2;
  sum1 = sum2 = 0;
  for (int i = 1; i <= n; i++) {
    cin >> f[i].remain;
    sum1 += f[i].remain;
  }
  for (int i = 1; i <= n; i++) {
    cin >> f[i].volume;
    sum2 += f[i].volume;
  }
  int x = sum1;
  int num = 0;
  sort(f + 1, f + 1 + n, cmp);
  while (1) {
    x -= f[++num].volume;
    if (x <= 0) break;
  }
  for (int i = 0; i <= 105; i++)
    for (int j = 0; j <= 10005; j++) dp[i][j] = -999999;
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = sum2; j >= f[i].volume; j--)
      for (int k = 1; k <= num; k++)
        dp[k][j] = max(dp[k][j], dp[k - 1][j - f[i].volume] + f[i].remain);
  }
  int ans = 0;
  for (int i = sum1; i <= sum2; i++) {
    ans = max(ans, dp[num][i]);
  }
  printf("%d %d\n", num, sum1 - ans);
}

```