#include <bits/stdc++.h>
using namespace std;
int n, f[105][10005];
int sum1 = 0, sum2 = 0;
int ans1 = 0, ans2 = -1;
struct node {
  int w;
  int v;
} a[105];
int cmp(node a, node b) { return a.v > b.v; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].w;
    sum1 += a[i].w;
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i].v;
    sum2 += a[i].v;
  }
  memset(f, 0xff, sizeof(f));
  f[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = n; j >= 1; j--) {
      for (int k = sum2; k >= a[i].v; k--) {
        if (f[j - 1][k - a[i].v] != -1) {
          f[j][k] = max(f[j][k], f[j - 1][k - a[i].v] + a[i].w);
        }
      }
    }
  }
  ans2 = -1;
  for (int j = 1; j <= n; j++) {
    for (int k = sum1; k <= sum2; k++) ans2 = max(ans2, f[j][k]);
    if (ans2 != -1) {
      cout << j << ' ' << sum1 - ans2 << endl;
      break;
    }
  }
}

### 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;
int n, f[105][10005];
int sum1 = 0, sum2 = 0;
int ans1 = 0, ans2 = -1;
struct node {
  int w;
  int v;
} a[105];
int cmp(node a, node b) { return a.v > b.v; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].w;
    sum1 += a[i].w;
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i].v;
    sum2 += a[i].v;
  }
  memset(f, 0xff, sizeof(f));
  f[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = n; j >= 1; j--) {
      for (int k = sum2; k >= a[i].v; k--) {
        if (f[j - 1][k - a[i].v] != -1) {
          f[j][k] = max(f[j][k], f[j - 1][k - a[i].v] + a[i].w);
        }
      }
    }
  }
  ans2 = -1;
  for (int j = 1; j <= n; j++) {
    for (int k = sum1; k <= sum2; k++) ans2 = max(ans2, f[j][k]);
    if (ans2 != -1) {
      cout << j << ' ' << sum1 - ans2 << endl;
      break;
    }
  }
}

```