#include <bits/stdc++.h>
using namespace std;
const int maxn = 105;
struct node {
  int a, b;
} num[maxn];
inline bool cmp(node x, node y) {
  if (x.b != y.b) return x.b > y.b;
  return x.a > y.a;
}
int f[maxn][maxn * maxn], n, m, res, numm, tot, sum;
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; i++)
    scanf("%d", &num[i].a), sum = sum + num[i].a;
  for (register int i = 1; i <= n; i++) scanf("%d", &num[i].b);
  sort(num + 1, num + n + 1, cmp);
  for (register int i = 1; i <= n; i++) {
    tot += num[i].b;
    if (tot >= sum) {
      numm = i;
      break;
    }
  }
  printf("%d ", numm);
  memset(f, -1, sizeof(f));
  f[0][0] = 0;
  for (register int i = 1; i <= n; i++)
    for (register int k = tot; k >= num[i].b; k--)
      for (register int j = 1; j <= numm; j++)
        if (f[j - 1][k - num[i].b] != -1)
          f[j][k] = max(f[j][k], f[j - 1][k - num[i].b] + num[i].a);
  for (register int i = sum; i <= tot; i++) res = max(res, f[numm][i]);
  printf("%d\n", sum - res);
  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 maxn = 105;
struct node {
  int a, b;
} num[maxn];
inline bool cmp(node x, node y) {
  if (x.b != y.b) return x.b > y.b;
  return x.a > y.a;
}
int f[maxn][maxn * maxn], n, m, res, numm, tot, sum;
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; i++)
    scanf("%d", &num[i].a), sum = sum + num[i].a;
  for (register int i = 1; i <= n; i++) scanf("%d", &num[i].b);
  sort(num + 1, num + n + 1, cmp);
  for (register int i = 1; i <= n; i++) {
    tot += num[i].b;
    if (tot >= sum) {
      numm = i;
      break;
    }
  }
  printf("%d ", numm);
  memset(f, -1, sizeof(f));
  f[0][0] = 0;
  for (register int i = 1; i <= n; i++)
    for (register int k = tot; k >= num[i].b; k--)
      for (register int j = 1; j <= numm; j++)
        if (f[j - 1][k - num[i].b] != -1)
          f[j][k] = max(f[j][k], f[j - 1][k - num[i].b] + num[i].a);
  for (register int i = sum; i <= tot; i++) res = max(res, f[numm][i]);
  printf("%d\n", sum - res);
  return 0;
}

```